As charge distribution is over a circle, potential due to charge over ring is 

$V=\frac{k}{r}.q_{\mathrm{total}}$

where, $q_{\mathrm{total} }={\int }_0^{2\pi }\frac{q{\sin }^2\theta }{\pi r}\left(rd\theta \right)$

$=\frac{q}{\pi }{\int }_0^{2\pi }\frac{1-\cos 2\theta }{2}.d\theta$

$=\frac{q}{\pi }{\left[\left(\frac{\theta }{2}-\frac{\sin 2\theta }{4}\right)\right]}_0^{2\pi }=q$

So, potential at centre of ring $=V$

$=\frac{k{q}_{\mathrm{total}}}{r}$

$\Rightarrow V=\frac{kq}{r}$

Also, work done in taking a charge $Q$ from centre of ring to infinity $=$ potential energy of system

$=V·Q=\frac{kqQ}{r}=\frac{qQ}{4\pi {\epsilon }_{0}r}$