The electronic configuration with maximum exchange energy will be:
A.$3d_{xy}^1\, 3d_{yz}^1 \,3d_{zx}^1\, 4{s^1}$
B.$3d_{xy}^1\,3d_{yz}^1\,3d_{zx}^1\,3d_{{x^2} - {y^2}}^1\,3d_{{z^2}}^1\,4{s^1}$
C.$3d_{xy}^2\,3d_{yz}^2\,3d_{zx}^2\,3d_{{x^2} - {y^2}}^2\,3d_{{z^2}}^1\,4{s^1}$
D.$3d_{xy}^2\,3d_{yz}^2\,3d_{zx}^2\,3d_{{x^2} - {y^2}}^2\,3d_{{z^2}}^2\,4{s^1}$

Hint: This question gives knowledge about the exchange energy. Exchange energy is the amount of energy released when two or more than two electrons having the same spin in the degenerate orbitals exchanges their spin.

Formula used: The formula used to determine the exchange energy is as follows:
$^n{C_2}$
Where $n$ is the number of electrons having the same spins and $C$ is used for combination.
The formula used to determine combination is as follows:
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where $n$ is the number of electrons having the same spins and $r$ is the total number of spin one orbital can have.

Complete step by step answer:
Exchange energy is the amount of energy released when two or more than two electrons having the same spin in the degenerate orbitals exchange their spin. The exchange energy is generally represented by $^n{C_2}$ .
Consider $3d_{xy}^1\,3d_{yz}^1\,3d_{zx}^1\,4{s^1}$ to determine the exchange energy.
In this electronic configuration there are four electrons present with the same spin. Therefore, we have to calculate a combination of $^4{C_2}$ to determine the exchange energy.
The formula used to determine the exchange energy is as follows:
${ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Substitute $n$ as $4$ and $r$as $2$ in the above formula as follows:
${ \Rightarrow ^4}{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}$
On simplifying, we get
${ \Rightarrow ^4}{C_2} = \dfrac{{4!}}{{2!\left( 2 \right)!}}$
On further simplifying, we get
${ \Rightarrow ^4}{C_2} = 6$
Therefore, option A has exchange energy as $6$.
Consider $3d_{xy}^1\,3d_{yz}^1\,3d_{zx}^1\,3d_{{x^2} - {y^2}}^1\,3d_{{z^2}}^1\,4{s^1}$ to determine the exchange energy.
In this electronic configuration there are six electrons present with the same spin. Therefore, we have to calculate a combination of $^6{C_2}$ to determine the exchange energy.
The formula used to determine the exchange energy is as follows:
${ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Substitute $n$ as $6$ and $r$ as B in the above formula as follows:
${ \Rightarrow ^6}{C_2} = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}$
On simplifying, we get
${ \Rightarrow ^6}{C_2} = \dfrac{{6!}}{{2!\left( 4 \right)!}}$
On further simplifying, we get
${ \Rightarrow ^6}{C_2} = 15$
Therefore, option B has exchange energy as $15$.
Consider $3d_{xy}^2\,3d_{yz}^2\,3d_{zx}^2\,3d_{{x^2} - {y^2}}^2\,3d_{{z^2}}^1\,4{s^1}$ to determine the exchange energy.
In this electronic configuration there are six electrons present with the same spin and four electrons with opposite spin. Therefore, we have to calculate a combination of $^6{C_2}$ in addition with $^4{C_2}$ to determine the exchange energy.
The formula used to determine the exchange energy is as follows:
${ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Substitute $n$ as $6$ and $r$ as $2$ and for opposite spin $n$ as $4$ and $r$ as $2$ in the above formula as follows:
${ \Rightarrow ^6}{C_2}{ + ^4}{C_2} = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}} + \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}$
On simplifying, we get
${ \Rightarrow ^6}{C_2}{ + ^4}{C_2} = \dfrac{{6!}}{{2!\left( 4 \right)!}} + \dfrac{{4!}}{{2!\left( 2 \right)!}}$
On further simplifying, we get
${ \Rightarrow ^6}{C_2}{ + ^4}{C_2} = 21$
Therefore, option C has exchange energy as $21$.
Consider $3d_{xy}^2\,3d_{yz}^2\,3d_{zx}^2\,3d_{{x^2} - {y^2}}^2\,3d_{{z^2}}^2\,4{s^1}$ to determine the exchange energy.
In this electronic configuration there are six electrons present with the same spin and five electrons with opposite spin. Therefore, we have to calculate a combination of $^6{C_2}$ in addition with $^5{C_2}$ to determine the exchange energy.
The formula used to determine the exchange energy is as follows:
${ \Rightarrow ^n}{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Substitute $n$ as $6$ and $r$ as $2$ and for opposite spin $n$ as $5$ and $r$ as $2$ in the above formula as follows:
${ \Rightarrow ^6}{C_2}{ + ^5}{C_2} = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}} + \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}}$
On simplifying, we get
${ \Rightarrow ^6}{C_2}{ + ^5}{C_2} = \dfrac{{6!}}{{2!\left( 4 \right)!}} + \dfrac{{4!}}{{2!\left( 3 \right)!}}$
On further simplifying, we get
${ \Rightarrow ^6}{C_2}{ + ^5}{C_2} = 25$
Therefore, option D has exchange energy as $25$.

Hence, option D is correct because it has maximum exchange energy.

Note: Always remember that the exchange energy is the amount of energy released when two or more than two electrons having the same spin in the degenerate orbitals exchange their spin. When antiparallel electrons are made to have parallel spin then the energy releases is exchange energy.