Two thin, conducting shells of radii R and 3R are shown in the figure. The outer shell carries a charge +Q and the inner shell is neutral. The inner shell is earthed with the help of a switch S.

When the switch is open, electric field inside the outer shell is zero,  potential at each point inside the outer shell is same, being equal to the potential on the surface of outer shell.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{inner}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Q}}{3\mathrm{R}} \\ {\mathrm{V}}_{\mathrm{outer}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Q}}{3\mathrm{R}} \end{gathered}$$

Option (A) is correct
When the switch is closed, the inner shell is grounded, its potential becomes zero.
Option (B) is correct

When S is closed, let q be the charge supplied to inner sphere by the earth

${\mathrm{V}}_{\mathrm{in}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{q}}{\mathrm{R}}+\frac{\mathrm{Q}}{3\mathrm{R}}\right]=0\Rightarrow \mathrm{q}=\frac{-\mathrm{Q}}{3}$

Option (C) is correct
Before closing the switch, capacitance,

${\text{C}}_{\text{1}}={\text{4πε}}_{\text{0}}\left(\text{3R}\right)={\text{12πε}}_{\text{0}}\text{R}$

After closing the switch, capacitance,

${\text{C}}_2=\frac{{\text{4πε}}_{\text{0}}{\left(\text{3R}\right)}^{\text{2}}}{\text{3R}-\text{R}}=18{\text{πε}}_{\text{0}}\text{R}$. Capacitance is increased.
Option (D) is correct.