The coercivity of a small magnet where the ferromagnet gets demagnetised is $3 \times {10^3}{\text{A}}{{\text{m}}^{ - 1}}$. The current required to be passed in a solenoid of length $10cm$ and number of turns $100$, so that the magnet gets demagnetized when inside the solenoid is:
(A) $3{\text{A}}$
(B) ${\text{6A}}$
(C) ${\text{30mA}}$
(D) ${\text{60mA}}$

Hint: To solve this question, we need to use the formula for the magnetic field inside a solenoid. Then, we have to use the relation between the magnetic field strength and the magnetic field intensity. The value of the coercivity given in the question refers to the intensity of the magnetic field.
Formula used: The formulae used in solving this question are given by
$B = {\mu _0}H$, here $B$ is the magnetic field strength, $H$ is the magnetic field intensity, and ${\mu _0}$ is the magnetic permeability of the free space.
$B = {\mu _0}nI$, here $B$ is the magnetic field strength inside a solenoid having $n$ turns per unit length, when a current of $I$ flows through it, and ${\mu _0}$ is the magnetic permeability of the free space.

Complete step-by-step solution:
We know that the coercivity of a magnet is equal to the magnetic field intensity necessary to demagnetize it. When the given small magnet is placed inside the solenoid, the intensity of the magnetic field generated due to the current flowing through the solenoid will demagnetize it. Let the current necessary to demagnetize the magnet be $I$. We know that the magnitude of the magnetic field strength generated inside a solenoid is given by
$B = {\mu _0}nI$ (1)
According to the question, the given solenoid has $100$ number of turns, and has a length of $10cm$. So the number of turns per unit length is given by
$n = \dfrac{N}{l}$
Substituting $N = 100$ and $l = 10cm = 0.1m$, we get
$n = \dfrac{{100}}{{0.1}}$
$ \Rightarrow n = 1000{m^{ - 1}}$ (2)
Substituting (2) in (1) we get
$B = 1000{\mu _0}I$ (3)
Now, we know that the intensity of magnetic field is related to the magnetic field strength by the relation
$B = {\mu _0}H$
$ \Rightarrow H = \dfrac{B}{{{\mu _0}}}$ (4)
Substituting (3) in (4) we get
$H = \dfrac{{1000{\mu _0}I}}{{{\mu _0}}}$
$ \Rightarrow H = 1000I$
According to the question, we have $H = 3 \times {10^3}{\text{A}}{{\text{m}}^{ - 1}}$. Substituting this above we get
$3 \times {10^3} = 1000I$
$ \Rightarrow I = 3A$
Thus, the current required to be passed through the solenoid is equal to $3{\text{A}}$.

Hence, the correct answer is option A.

Note: Do not forget to convert the length given in the question into the SI unit, while calculating the number of turns of the solenoid per unit length. Also, the coercivity refers to the intensity of the magnetic field, not to the magnetic field strength.