From figure, $\mathrm{d} \ell=\mathrm{Rd} \theta$

Charge on $d \ell=\lambda \mathrm{Rd} \theta$

where $\lambda=$ linear charge density.

Electric field at centre due to $d \ell$

$\mathrm{dE}=\mathrm{k} \cdot \frac{\lambda \mathrm{Rd} \theta}{\mathrm{R}^{2}}$

We need to consider only the component $dE$ $\cos \theta,$ as the component $dE$ sin $\theta$ will cancel out.

$\therefore  $ Total field at centre $=2 \int_{0}^{\pi / 2} \mathrm{dE}\, \cos \theta$

$=2 \int_{0}^{\pi / 2} \frac{\mathrm{k} \lambda \mathrm{R}\, \cos \theta}{\mathrm{R}^{2}} \mathrm{d} \theta=\frac{2 \mathrm{k} \lambda}{\mathrm{R}} \int_{0}^{\pi / 2} \cos\, \theta \mathrm{d} \theta$

$=\frac{9}{2 \pi^{2} \epsilon_{0} R^{2}} \quad\left(\text { since } \lambda=\frac{q}{\pi R}\right)$