Answer
See the detailed answer below.
Work Step by Step
First of all, we need to draw the electric field directions on each point.
$\textbf{At point 1:}$
From the geometry of the first figure below, we can see that
$\theta_1=\theta_2$,
$r_{11}=r_{21}=\sqrt{a^2+(2a)^2}=\sqrt{5}\;a$,
$\sin\theta=\dfrac{a}{\sqrt{5}\;a}=\dfrac{1}{\sqrt{5}}$,
$\cos\theta=\dfrac{2a}{\sqrt{5}\;a}=\dfrac{2}{\sqrt{5}}$,
Hence,
$$E_{1\rightarrow1}=\dfrac{k|q_1|}{r_{11}^2}(\sin\theta\;\hat i+\cos\theta\;\hat j)$$
Plug the known;
$$E_{1\rightarrow1}=\dfrac{2kq}{5a^2}(\dfrac{1}{\sqrt5}\;\hat i+\dfrac{2}{\sqrt{5}}\;\hat j)\tag 1$$
And
$$E_{2\rightarrow1}=\dfrac{k|q_2|}{r_{21}^2}(\sin\theta\;\hat i-\cos\theta\;\hat j)$$
Plug the known;
$$E_{2\rightarrow1}=\dfrac{kq}{5a^2}(\dfrac{1}{\sqrt5}\;\hat i-\dfrac{2}{\sqrt{5}}\;\hat j)\tag 2$$
So the net electric field at point 1 is given by
$$E_{1,net}=E_{1\rightarrow1}+E_{2\rightarrow1}$$
Plug from (1) and (2),
$$E_{1,net}=\dfrac{2kq}{5a^2}(\dfrac{1}{\sqrt5}\;\hat i+\dfrac{2}{\sqrt{5}}\;\hat j)+\dfrac{kq}{5a^2}(\dfrac{1}{\sqrt5}\;\hat i-\dfrac{2}{\sqrt{5}}\;\hat j)$$
$$E_{1,net}=\dfrac{ kq}{5\sqrt{5}a^2}\left[ (2\;\hat i+4\;\hat j)+ (1\;\hat i-2\;\hat j)\right]$$
$$\boxed{E_{1,net}=\dfrac{ kq}{5\sqrt{5}a^2}\left( 3\;\hat i+2\;\hat j \right)}$$
$\textbf{At point 2:}$
$$E_{1\rightarrow2}=-\dfrac{k|q_1|}{r_{12}^2}\;\hat i$$
Plug the known;
$$E_{1\rightarrow2}=-\dfrac{2kq}{a^2}\;\hat i\tag 3$$
$$E_{2\rightarrow2}= \dfrac{k|q_2|}{r_{22}^2}\;\hat i$$
Plug the known;
$$E_{2\rightarrow2}= \dfrac{ kq}{9a^2}\;\hat i\tag 4$$
So the net electric field at point 2 is given by
$$E_{2,net}=E_{1\rightarrow2}+E_{2\rightarrow2}$$
Plug from (3) and (4),
$$E_{2,net}=-\dfrac{2kq}{a^2}\;\hat i+ \dfrac{ kq}{9a^2}\;\hat i=\dfrac{-18kq+kq}{9a^2}\;\hat i$$
$$\boxed{E_{2,net} =\dfrac{-17kq}{9a^2}\;\hat i}$$
$\textbf{At point 3:}$
$$E_{1\rightarrow3}=\dfrac{k|q_1|}{r_{13}^2}\;\hat i$$
Plug the known;
$$E_{1\rightarrow3}=\dfrac{2kq}{9a^2}\;\hat i\tag 7$$
$$E_{2\rightarrow3}=- \dfrac{k|q_2|}{r_{23}^2}\;\hat i$$
Plug the known;
$$E_{2\rightarrow3}= -\dfrac{ kq}{a^2}\;\hat i\tag 8$$
So the net electric field at point 3 is given by
$$E_{3,net}=E_{1\rightarrow3}+E_{2\rightarrow3}$$
Plug from (7) and (8),
$$E_{3,net}=\dfrac{2kq}{9a^2}\;\hat i -\dfrac{ kq}{a^2}\;\hat i=\dfrac{2kq-9kq}{9a^2}\;\hat i$$
$$\boxed{E_{3,net} =\dfrac{ -7kq}{9a^2}\;\hat i}$$
$\textbf{At point 4:}$
From the geometry of the first figure below, we can see that
$\theta_1=\theta_2$,
$r_{14}=r_{24}=\sqrt{a^2+(2a)^2}=\sqrt{5}\;a$,
$\sin\theta=\dfrac{a}{\sqrt{5}\;a}=\dfrac{1}{\sqrt{5}}$,
$\cos\theta=\dfrac{2a}{\sqrt{5}\;a}=\dfrac{2}{\sqrt{5}}$,
Hence,
$$E_{1\rightarrow4}=\dfrac{k|q_1|}{r_{14}^2}(\sin\theta\;\hat i-\cos\theta\;\hat j)$$
Plug the known;
$$E_{1\rightarrow4}=\dfrac{2kq}{5a^2}(\dfrac{1}{\sqrt5}\;\hat i-\dfrac{2}{\sqrt{5}}\;\hat j)\tag 5$$
And
$$E_{2\rightarrow4}=\dfrac{k|q_2|}{r_{24}^2}(\sin\theta\;\hat i+\cos\theta\;\hat j)$$
Plug the known;
$$E_{2\rightarrow4}=\dfrac{kq}{5a^2}(\dfrac{1}{\sqrt5}\;\hat i+\dfrac{2}{\sqrt{5}}\;\hat j)\tag 6$$
So the net electric field at point 4 is given by
$$E_{4,net}=E_{1\rightarrow4}+E_{2\rightarrow4}$$
Plug from (5) and (6),
$$E_{4,net}=\dfrac{2kq}{5a^2}(\dfrac{1}{\sqrt5}\;\hat i-\dfrac{2}{\sqrt{5}}\;\hat j)+\dfrac{kq}{5a^2}(\dfrac{1}{\sqrt5}\;\hat i+\dfrac{2}{\sqrt{5}}\;\hat j)$$
$$E_{4,net}=\dfrac{ kq}{5\sqrt{5}a^2}\left[ (2\;\hat i-4\;\hat j)+ (1\;\hat i+2\;\hat j)\right]$$
$$\boxed{E_{4,net}=\dfrac{ kq}{5\sqrt{5}a^2}\left( 3\;\hat i-2\;\hat j \right)}$$
