The figure shows equipotential surfaces concentric at O. The magnitude of electric field at a distance r metres from O is

Potential due to a point charge ‘q’ at a distance ‘r’ from it is given by

$V=\frac{1}{4\pi {\in }_0}\frac{q}{r}$

For $60V$ equipotential surface

$60=\frac{1}{4\pi {\in }_0}.\frac{q}{10\times {10}^{-2}}\Rightarrow \frac{q}{4\pi {\in }_0}=6$--(1)

So electric field at a distance ‘r’ from charge ‘q’ is.

$E=\frac{1}{4\pi {\in }_0}\frac{q}{r^2}=\frac{6}{r^2}V{m}^{-1}$