The three charges (q)/(2), q and (q)/(2) are placed at the corners...

${E}_{A} = \dfrac{k\left(\dfrac{q}{2}\right)}{{a}^{2}} = \dfrac{kq}{2{a}^{2}}$, ${E}_{C} = \dfrac{k\left(\dfrac{q}{2}\right)}{{a}^{2}} = \dfrac{kq}{2{a}^{2}}$ and ${E}_{D} = \dfrac{k\left(q\right)}{{\left(\sqrt{2}a\right)}^{2}} = \dfrac{kq}{2{a}^{2}}$

As we can see from diagram, ${E}_{C} \text{ and } {E}_{A}$ are perpendicular.

Therefore, net electric field will be,

$= \dfrac{kq}{2{a}^{2}} + \sqrt{{\left(\dfrac{kq}{2{a}^{2}}\right)}^{2} + {\left(\dfrac{kq}{2{a}^{2}}\right)}^{2}} = \dfrac{kq}{2{a}^{2}} + \dfrac{kq}{\sqrt{2}{a}^{2}} = \dfrac{q}{4\pi{ϵ}_{0}{a}^{2}}\left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\right)$