Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
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
$$E_1=-\dfrac{k|q_1|}{r_1^2}\;\hat i$$
Plug the known;
$$E_1=-\dfrac{(9.0\times 10^{ 9})(10\times 10^{-9})}{0.04^2}\;\hat i$$
$$E_1=({\bf -56,250}\;\hat i )\;{\rm N/C}\tag 1$$
$$E_2=\dfrac{k|q_2|}{r_2^2}(-\cos\theta\;\hat i-\sin\theta\;\hat j)$$
where $\sin\theta= 0.02/r_2$, $\cos\theta=0.04/r_2$.
Thus,
$$E_2=\dfrac{k|q_2|}{r_2^3}(-0.04 \;\hat i-0.02\;\hat j)$$
where $r_2=\sqrt{0.02^2+0.04^2}$
$$E_2=\dfrac{(9.0\times 10^{ 9})(10\times 10^{-9})}{(0.02^2+0.04^2)^{3/2}}(-0.04 \;\hat i-0.02\;\hat j)$$
$$E_2=({\bf -40,249}\;\hat i-{\bf 20,125}\;\hat j)\;\rm N/C\tag 2$$
$$E_3= \dfrac{kq_3}{r_3^2}\;\hat j$$
Plug the known;
$$E_3= \dfrac{(9.0\times 10^{ 9})(5\times 10^{-9})}{0.02^2}\;\hat j$$
$$E_3=({\bf 112,500}\;\hat j )\;{\rm N/C}\tag 3$$
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),
$$E_{net}=[ -56,250\;\hat i]+[ -40,249\;\hat i- 20,125\;\hat j]+[112,500\;\hat j] $$
$$E_{net}= (\color{red}{\bf -96,499}\;\hat i +\color{red}{\bf 92,375}\;\hat j)\;\rm N/C $$
$$\color{blue}{\bf [b]}$$
The net electric field as a magnitude is given by
$$E_{net}=\sqrt{E_x^2+E_y^2}$$
$$E_{net}=\sqrt{96,499^2+92,375^2}$$
$$E_{net}= \color{red}{\bf 133,586}\;\rm N/C$$
And since the $x$-component is negative and the $y$-component is positive, the angle of the net electric field is in the second quadrant.
Hence, its angle is given by
$$\theta_{net}=180^\circ-\tan^{-1}\left[ \dfrac{E_y}{E_x}\right]=180^\circ-\tan^{-1}\left[ \dfrac{92,375}{96,499}\right] $$
$$\theta_{net}=\color{red}{\bf 136.3}^\circ\tag{CCW}$$
Its direction is 136.3$^\circ$ counterclockwise from the $+x$-direction.
