A solid sphere of radius R is charged uniformly. At a point P, a distance r from the surface, the electrostatic potential is observed to be half of the potential at the centre. Then

Since, $V=\frac{kQ}{2R}(3R^2-r^2)$

So, at r = 0, potential at the centre is $V_c=\frac{3kQ}{2R}$

As per the question we require a point where  $V=\frac{V_c}{2}=\frac{3kQ}{4R}$

This point cannot lie inside the sphere where $V\ge \frac{kQ}{R}$

Let the point lie outside the sphere, at a distance r+R from the centre. Then

$V=\frac{kQ}{(R+r)}=\frac{3kQ}{4R}$

$\Rightarrow r=\frac{R}{3}$