Potential due to point charge $q$ at distance $r$ is given by $v=\frac{K q}{r}$
For $60\, V$ equipotential surface,
$60=\frac{k q}{r}, $ here $ r =10\, cm =10 \times 10^{-2} m $
$60 =\frac{k q}{10 \times 10^{-2}} $
$\therefore k q=60 \times 10 \times 10^{-2}=6\,\,\,...(i)$
The electric field at distance $r$ from charge $q$,
$E=\frac{k q}{r^{2}}$
Substitution from Eq. (i) gives
$E=\frac{6}{r^{2}} Vm ^{-1} $
