The semicircular non – conducting ring of radius R containing +ve and negative charge as shown are joined in two perpendicular planes. The magnitude of linear charge densities is uniform and is λ . The electric field at O will be

The electric field because of one semi circular charged where

$\mathrm{E}=\frac{2\mathrm{K}\lambda }{\mathrm{R}}$ along the axis of semi-circular ring.

As the two  semi circular rings are in perpendicular planes, so the resultant electric field will be

$\left|{\overrightarrow{\mathrm{E}}}_{\mathrm{net}}\right|=\sqrt{2}.\frac{2\mathrm{K\lambda }}{\mathrm{R}}=2\sqrt{2}\times \frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{\lambda }}{\mathrm{R}}=\frac{\mathrm{\lambda }}{\sqrt{2}{\mathrm{\pi ϵ}}_0\mathrm{R}}$