(a) We know that intensity is negative gradient of potential,
$ie,\ m\ I = - (dV/dr)$ and as here $I = - (k/r)$, so
$\frac{dV}{dr} = \frac{k}{r},\ ie,\ \int_{0}^{v}{dV} = k\int_{r_{0}}^{r}\frac{dr}{r}$
or $V - V_{0} = k\log\frac{r}{r_{0}}\ \ \ so,V = k\log{\frac{r}{r_{0}} + V_{0}}$