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;
$$E_1=\dfrac{(9.0\times 10^{ 9})(5\times 10^{-9})}{0.02^2}\;\hat i$$
$$E_1=({\bf 112,500}\;\hat i )\;{\rm N/C}\tag 1$$
$$E_2=\dfrac{k|q_2|}{r_2^2}(-\sin\theta\;\hat i+\sin\theta\;\hat j)$$
where $\sin\theta= 0.02/r_2$, $\cos\theta=0.04/r_2$.
Thus,
$$E_2=\dfrac{k|q_2|}{r_2^3}(-0.02 \;\hat i+0.04\;\hat j)$$
where $r_2=\sqrt{0.02^2+0.04^2}$
$$E_2=\dfrac{(9.0\times 10^{ 9})(5\times 10^{-9})}{(0.02^2+0.04^2)^{3/2}}(-0.02 \;\hat i+0.04\;\hat j)$$
$$E_2=({\bf -10,062}\;\hat i+{\bf 20,124}\;\hat j)\;\rm N/C\tag 2$$
$$E_3=-\dfrac{kq_3}{r_3^2}\;\hat j$$
Plug the known;
$$E_3=-\dfrac{(9.0\times 10^{ 9})(10\times 10^{-9})}{0.04^2}\;\hat j$$
$$E_3=({\bf -56,250
}\;\hat j )\;{\rm N/C}\tag 3$$
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),
$$E_{net}=[112,500\;\hat i]+[ -10,062\;\hat i+ 20,124\;\hat j]+[-56,250\;\hat j] $$
$$E_{net}= (\color{red}{\bf 102,438}\;\hat i -\color{red}{\bf 36,126}\;\hat j)\;\rm N/C $$
$$\color{blue}{\bf [b]}$$
The net electric field as a magnitude is given by
$$E_{net}=\sqrt{E_x^2+E_y^2}$$
$$E_{net}=\sqrt{102,438^2+36,126^2}$$
$$E_{net}= \color{red}{\bf 108,621}\;\rm N/C$$
And since the $x$-component is positive and the $y$-component is negative, the angle of the net electric field is in the fourth quadrant.
Hence, its angle is given by
$$\theta_{net}=\tan^{-1}\left[ \dfrac{E_y}{E_x}\right]=\tan^{-1}\left[ \dfrac{36,126}{102,438}\right]$$
$$\theta_{net}=\color{red}{\bf19.4}^\circ\tag{CW}$$
Its direction is 19.4$^\circ$ clockwise from the $+x$-direction.
