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;
$$E_1= \dfrac{(9.0\times 10^{ 9})(5\times 10^{-9})}{0.05^2}\;\hat i$$
$$E_1=({\bf 18,000}\;\hat i )\;{\rm N/C}\tag 1$$
$$E_2=\dfrac{k|q_2|}{r_2^2}(-\cos\theta\;\hat i-\sin\theta\;\hat j)$$
where $\sin\theta= 0.03/r_2$, $\cos\theta=0.05/r_2$.
Thus,
$$E_2=\dfrac{k|q_2|}{r_2^3}(-0.05 \;\hat i-0.03\;\hat j)$$
where $r_2=\sqrt{0.03^2+0.05^2}$
$$E_2=\dfrac{(9.0\times 10^{ 9})(10\times 10^{-9})}{(0.03^2+0.05^2)^{3/2}}(-0.05 \;\hat i-0.03\;\hat j)$$
$$E_2=({\bf -22,698}\;\hat i-{\bf 13,619}\;\hat j)\;\rm N/C\tag 2$$
$$E_3= \dfrac{kq_3}{r_3^2}\;\hat j$$
Plug the known;
$$E_3= \dfrac{(9.0\times 10^{ 9})(10\times 10^{-9})}{0.03^2}\;\hat j$$
$$E_3=({\bf 100,000}\;\hat j )\;{\rm N/C}\tag 3$$
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),
$$E_{net}=[ 18,000\;\hat i]+[ -22,698\;\hat i- 13,619\;\hat j]+[100,000\;\hat j] $$
$$E_{net}= (\color{red}{\bf -4,698}\;\hat i +\color{red}{\bf 86,381}\;\hat j)\;\rm N/C $$
$$\color{blue}{\bf [b]}$$
The net electric field as a magnitude is given by
$$E_{net}=\sqrt{E_x^2+E_y^2}$$
$$E_{net}=\sqrt{4,698^2+86,381^2}$$
$$E_{net}= \color{red}{\bf 86,509}\;\rm N/C$$
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
$$\theta_{net}=180^\circ-\tan^{-1}\left[ \dfrac{E_y}{E_x}\right]=180^\circ-\tan^{-1}\left[ \dfrac{86,381}{4,698}\right] $$
$$\theta_{net}=\color{red}{\bf 93.1}^\circ\tag{CCW}$$
Its direction is 93.1$^\circ$ counterclockwise from the $+x$-direction.
