Two concentric conducting spherical shells of radius ${{r}_{1}}$ and ${{r}_{2}}$ $\left( {{r}_{2}}\rangle {{r}_{1}} \right)$. Outer shell is given a charge Q and the inner is earthed. Charge on the inner sphere will be
$\begin{align}
  & \left( 1 \right)\dfrac{-Q{{r}_{2}}}{{{r}_{1}}} \\
 & \left( 2 \right)\dfrac{-Q{{r}_{1}}}{{{r}_{2}}} \\
 & \left( 3 \right)Zero \\
 & \left( 4 \right)\dfrac{-Qr_{1}^{2}}{r_{2}^{2}} \\
\end{align}$

Hint: When the inner sphere is earthed, then the charge can flow to the ground until the potential reaches to zero. Hence, here the total potential is equal to zero. Thus first calculate the potential ${{V}_{1}}$ and then calculate the second potential ${{V}_{2}}$. Since the sum of the two potentials is zero, equate the sum of potential equals to zero. Then by rearranging them we will get the charge at the inner sphere.

Formula used:
Potential at appoint is given by,
$V=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{q}{r}$
where, $\dfrac{1}{4\pi {{\in }_{0}}}$is constant.
$\Rightarrow $ $k=\dfrac{1}{4\pi {{\in }_{0}}}$and its value is $9\times {{10}^{9}}$
${{\in }_{0}}$ is the permittivity of free space.
q is the charge
and the r is the radius of the spherical shell.

Complete step by step answer:
Let the charge in the inner sphere be ‘q’.
When the inner sphere is earthed, then the charge can flow to the ground until the potential reaches zero.
Hence here the total potential must be zero.
Potential ${{V}_{1}}=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{q}{{{r}_{1}}}$
Here the term$\dfrac{1}{4\pi {{\in }_{0}}}$ is a constant.
$\Rightarrow $ $k=\dfrac{1}{4\pi {{\in }_{0}}}$
Hence potential ${{V}_{1}}=\dfrac{kq}{{{r}_{1}}}$
Thus potential, ${{V}_{2}}=\dfrac{1}{4\pi {{\in }_{0}}}\dfrac{Q}{{{r}_{2}}}$
$\Rightarrow {{V}_{2}}=\dfrac{kQ}{{{r}_{2}}}$
where, $k=\dfrac{1}{4\pi {{\in }_{0}}}$
Since the total potential here is zero.
$\Rightarrow {{V}_{1}}+{{V}_{2}}=0$
Thus by substituting the value of ${{V}_{1}}$and ${{V}_{2}}$ in the above equation we get,
$\dfrac{kq}{{{r}_{1}}}+\dfrac{kQ}{{{r}_{2}}}=0$
$\Rightarrow \dfrac{kq}{{{r}_{1}}}=-\dfrac{kQ}{{{r}_{2}}}$
$\therefore\dfrac{q}{{{r}_{1}}}=-\dfrac{Q}{{{r}_{2}}}$
Therefore, $q=-Q\dfrac{{{r}_{1}}}{{{r}_{2}}}$.
Thus option (B) is correct.

Note:
When the inner sphere is earthed, then the charge can flow to the ground until the potential reaches zero. That is the sum of potential is zero here. While calculating the first potential consider the inner radius and when calculating the second potential consider the outer radius. That is earth a charge means the potential gradually reduces to zero.