A solid conducting sphere of radius ‘a’ is surrounded by a thin uncharged concentric conducting shell of radius 2a. A point charge q is placed at a distance 4a from common centre of conducting sphere and shell. The inner sphere is then grounded. 

The potential of outer shell is

The inner sphere is grounded, hence its potential is zero. 

The net charge on isolated outer sphere is zero. 

Let the charge on inner sphere be q’.

$\therefore$Potential at centre of inner sphere is

$=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^{\mathrm{\prime }}}{\mathrm{a}}+0+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{4\mathrm{a}}=0\therefore {\mathrm{q}}^{\mathrm{\prime }}=-\frac{\mathrm{q}}{4}$

The region in between conducting sphere and shell in shielded from charges on and outside surface of shell. Hence charge distribution on surface of sphere and inner surface of shell is uniform.

The distribution of induced charge on outer surface of shell depends only on charge q, hence is non-uniform.

The electric field at B is $=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{\mathrm{q}}{4{\mathrm{x}}^2}$ towards left.

$\therefore {\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{A}}=-{\int }_{2\mathrm{a}}^{\mathrm{a}} \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{4{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{32{\mathrm{\pi \epsilon }}_0}\cdot \frac{\mathrm{q}}{\mathrm{a}}$