a b c d

c

$\text{Potential at the centre } V_c = \frac{3}{2} \frac{kQ}{R}$ $\text{Potential at the distance x from centre } V = \frac{kQ}{x}$ $ \text{It is given that } V = \frac{V_c}{4}$ $\rightarrow \frac{kQ}{x} = \frac{1}{4} \frac{3kQ}{2R}$ $\Rightarrow x = \frac{8R}{3}$ $\text{Distance from Surface is } =\frac{8R}{3} - R = \frac{5R}{3}$