The electric potential at the centre of the complete disc, $V_1=\frac{\sigma R}{2{\epsilon }_0}$

Let us assume that the potential at the circumference of the smaller disc is $V_2$

$r=\frac{2R}{2}\mathit{cos}\left(\frac{\theta }{2}\right)$

$\mathrm{d}r=-\frac{2R}{2}\mathit{sin}\left(\frac{\theta }{2}\right)\left(\frac{d\theta }{2}\right)$

$\int \mathrm{d}{V}_2={\int }_{\theta = 0}^{\pi }\frac{\sigma \times \left(r\theta \right)\left(\mathrm{d}r\right)}{4\pi {\epsilon }_{0}r}=\frac{\sigma R}{2\pi {\epsilon }_0}=V_2$

$\therefore$ Net potential at centre $\mathrm{C}$ is,

$V=\left(V_1–V_2\right)=\frac{\sigma R\left(\pi -1\right)}{2\pi {\epsilon }_0}$