Solve this:

Q. Consider a sphere of radius R having charge q uniformly distributed inside it. At what minimum distance from its surface the electric potential is half of the electric potential at its centre? 
$$\begin{gathered} \left(1\right) \mathrm{R} \left(2\right) \frac{\mathrm{R}}{2} \\ \left(3\right) \frac{4\mathrm{R}}{3} \left(4\right) \frac{\mathrm{R}}{3} \end{gathered}$$

we know potential outside the sphere is V1 = kQ/x     where x is the distance from the center x>r

 and potential inside the sphere V1 = kQ/2r [3-x2/r2]

 for center put x=0  Vo = 3kQ/2r

condition given is V1 = Vo/2

                      kQ/x  = (3kQ/2r)/2

                         x=4r/3

 distance from the surface is  =x-r

                                         = 4r/3 -r =r/3
Regards