$E_A=\frac{k\left({\displaystyle \frac{q}{2}}\right)}{a^2}=\frac{kq}{2a^2}$,  $E_C=\frac{k\left({\displaystyle \frac{q}{2}}\right)}{a^2}=\frac{kq}{2a^2}$  and  $E_D=\frac{k\left({\displaystyle q}\right)}{{\left(\sqrt{2}a\right)}^2}=\frac{kq}{2a^2}$

As we can see from diagram, $E_C \& E_A$ are perpendicular.

Therefore, net electric field will be,

$$\begin{gathered} =\frac{kq}{2a^2}+\sqrt{{\left(\frac{kq}{2a^2}\right)}^2+{\left(\frac{kq}{2a^2}\right)}^2}=\frac{kq}{2a^2}+\frac{kq}{\sqrt{2}{a}^2} \\ =\frac{q}{4\pi {ϵ}_0{a}^2}\left(\frac{1}{\sqrt{2}}+\frac{1}{2}\right) \end{gathered}$$