Answer
See the detailed answer below.
Work Step by Step
First, we need to draw the electric field direction at the black dot, as seen below.
Thus, from the figure below, we can see that
$$E_1=-\dfrac{k|q_1|}{r_1^2}\;\hat j$$
Plug the known;
$$E_1=-\dfrac{kQ}{L^2}\;\hat j \tag 1$$
$$E_2=\dfrac{k|q_2|}{r_2^2}(\cos\theta\;\hat i+\sin\theta\;\hat j)$$
where $\sin\theta= L/r_2$, $\cos\theta=L/r_2$.
Thus,
$$E_2=\dfrac{k|q_2|}{r_2^3}(L\;\hat i+L\;\hat j)$$
where $r_2=\sqrt{L^2+L^2}=\sqrt{2}L$
$$E_2=\dfrac{4kQL}{(\sqrt{2}L)^3}(1\;\hat i+1\;\hat j)$$
$$E_2=\dfrac{4kQ }{2^{3/2}{L^2}} \;\hat i+\dfrac{4kQ }{2^{3/2}{L^2}}\;\hat j \tag 2$$
$$E_3= -\dfrac{kq_3}{r_3^2}\;\hat i$$
Plug the known;
$$E_3= -\dfrac{kQ}{L^2}\;\hat i\tag 3$$
Therefore, the net electric field in a component form is given by
$$E_{net}=E_1+E_2+E_3$$
Plugging from above (1), (2), (3),
$$E_{net}= \left[-\dfrac{kQ}{L^2}\;\hat j\right]+\left[\dfrac{4kQ }{2^{3/2}{L^2}} \;\hat i+\dfrac{4kQ }{2^{3/2}{L^2}}\;\hat j \right]+\left[-\dfrac{kQ}{L^2}\;\hat i\right]$$
$$E_{net}= \left[\dfrac{4kQ }{2^{3/2}{L^2}} -\dfrac{kQ}{L^2} \right] \;\hat i+\left[ \dfrac{4kQ }{2^{3/2}{L^2}}-\dfrac{kQ}{L^2}\;\right]\hat j$$
$$E_{net}= \left[\dfrac{4kQ -2^{3/2}kQ}{2^{3/2}{L^2}} \right] \;\hat i+\left[ \dfrac{4kQ -2^{3/2}kQ}{2^{3/2}{L^2}} \;\right]\hat j$$
$$E_{net}= \left[\dfrac{kQ( 4-2\sqrt2)}{2\sqrt{2}{L^2}} \right] \;\hat i+\left[ \dfrac{kQ( 4-2\sqrt2)}{2\sqrt{2}{L^2}} \;\right]\hat j$$
$$\boxed{E_{net}= \left[\dfrac{kQ( \sqrt{2}-1)}{{L^2}} \right] \;\hat i+\left[ \dfrac{kQ( \sqrt{2}-1)}{{L^2}} \;\right]\hat j}$$