The charge per unit length of the four quadrant of the ring is 2$\mathrm{\lambda }$, – 2$\mathrm{\lambda }$ , $\mathrm{\lambda }$ and -$\mathrm{\lambda }$ respectively. The electric field at the centre is

$\begin{array}{l}\mathrm{\ell }=\mathrm{R\theta }\Rightarrow \mathrm{d\ell }=\mathrm{Rd\theta }\\ {\mathrm{E}}_{\mathrm{x}}=\int \mathrm{dEcos}\mathrm{\theta }\text{; }\\ {\mathrm{E}}_{\mathrm{x}}=\int \frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{dq}}{{\mathrm{R}}^2}\cos \mathrm{\theta }\\ =\int \frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{\lambda d\ell }}{{\mathrm{R}}^2}\cos \mathrm{\theta }\\ ={\int }_0^{\mathrm{\pi }/2} \frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{\lambda cos}\mathrm{\theta d\theta }}{\mathrm{R}}=\frac{\mathrm{\lambda }}{4\mathrm{\pi }{\in }_0\mathrm{R}}\end{array}$

${\mathrm{E}}_{\mathrm{y}}={\int }_0^{\mathrm{\pi }/2} \frac{1}{4{\mathrm{\pi E}}_0}\frac{\mathrm{\lambda }}{\mathrm{R}}\sin \mathrm{\theta d\theta }=-\frac{1}{4{\mathrm{\pi E}}_0}\frac{\mathrm{\lambda }}{\mathrm{R}}$

$\begin{array}{l}{\mathrm{E}}_{\text{net }}=2\mathrm{Esin}\mathrm{\theta }/2\Rightarrow {\mathrm{E}}_{\text{net }}=\frac{2\sqrt{2\mathrm{\lambda }}}{4\mathrm{\pi }{\in }_0\mathrm{R}}\frac{1}{\sqrt{2}}\\ =\frac{\mathrm{\lambda }}{2\mathrm{\pi }{\in }_0\mathrm{R}}(-\widehat{\mathrm{i}})\Rightarrow {\mathrm{E}}_{\text{net }}=-\frac{\mathrm{\lambda }}{2\mathrm{\pi }{\in }_0\mathrm{R}}\widehat{\mathrm{i}}\end{array}$