$V=\frac{k}{r} \cdot q_{ total }$

$\text { where, }$

$q_{\text {total }} =\int \limits_{0}^{2 \pi} \frac{q \sin ^{2} \theta}{\pi r}(r d \theta)$

$=\frac{q}{\pi} \int \limits_{0}^{2 \pi} \frac{1-\cos 2 \theta}{2} \cdot d \theta$

$=\frac{q}{\pi}\left[\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)\right]_{0}^{2 \pi}=q$

So, potential at centre of ring $=V$

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

$\Rightarrow \quad V =\frac{k q}{r}$

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

$=V \cdot Q=\frac{k q Q}{r}=\frac{q Q}{4 \pi \varepsilon_{0} r}$