d
Given: Length of wire \(L=20 \,\mathrm{cm}\) charge \(Q=10^{3} \varepsilon_{0}\)

We know, electric field at the centre of the semicircular arc

\({E=\frac{2 K \lambda}{r}}\)

\({\text{ or, }}\quad E = \frac{{2K\left( {\frac{{2Q}}{{\pi r}}} \right)}}{r}\left[ {As\,\lambda  = \frac{{2Q}}{{\pi r}}} \right]\)

\(=\frac{4 K Q}{\pi r^{2}}=\frac{4 K Q \pi^{2}}{\pi L^{2}}=\frac{4 \pi K Q}{L^{2}}=25 \times 10^{3} \,N / C \hat{i}\)