$\text{Potential at the centre } V_c = \frac{3}{2} \frac{kQ}{R}$

$\text{Potential at the distance x from centre } V = \frac{kQ}{x}$

$ \text{It is given that } V = \frac{V_c}{4}$

$\rightarrow \frac{kQ}{x} = \frac{1}{4} \frac{3kQ}{2R}$

$\Rightarrow x = \frac{8R}{3}$

$\text{Distance from Surface is } =\frac{8R}{3} - R = \frac{5R}{3}$