Answer
$10\;\rm kN/C$
$-11.12^\circ$
Work Step by Step
The electric field direction from charge 1 is away from them while from charge 2 is toward the charge since it is a negative charge.
The upper charge exerts on the dot an electric force that is rightward and has no $ y$-component.
Thus, its angle with respect to $+x$-direction is zero.
$$\theta_1=\bf 0^\circ$$
The other charge exerting an electric force on the dot is at an angle $\theta_2$ counterclockwise with respect to $+x$-direction.
From the geometry of the figure below, we can see that
$$\theta_2=270^\circ-\phi $$
where $\phi=90^\circ-\theta$, so
$$\theta_2=270^\circ-(90^\circ-\theta)=180^\circ+\theta $$
Noting that,
$$\tan\theta=\dfrac{10}{5}=2 \rightarrow \theta = 63.43^\circ$$
Thus,
$$\theta_2=180^\circ+63.43^\circ=\bf 243.43^\circ$$
Thus, the net electric field in $x$-direction is given by
$$\sum E_x=\dfrac{kq_1}{r_1^2}\cos\theta_1 +\dfrac{kq_2}{r_2^2}\cos\theta_2$$
Noting that $q_1=q_2=q$,
$$\sum E_x=kq\left[ \dfrac{\cos\theta_1 }{r_1^2}+\dfrac{\cos\theta_2}{r_2^2}\right]$$
Plugging the known,
$$\sum E_x=(8.99\times 10^9)(3\times 10^{-9})\left[ \dfrac{\cos0^\circ}{0.05^2}+\dfrac{\cos 243.43^\circ }{(0.05^2+0.1^2)}\right]$$
$$\sum E_x=\bf 9823\;\rm N/C$$
By the same approach, the net electric field in $y$-direction is given by
$$\sum E_y=kq\left[ \dfrac{\sin\theta_1 }{r_1^2}+\dfrac{\sin\theta_2}{r_2^2}\right]$$
$$\sum E_y=(8.99\times 10^9)(3\times 10^{-9})\left[ \dfrac{\sin0^\circ}{0.05^2}+\dfrac{\sin 243.43^\circ }{(0.05^2+0.1^2)}\right]$$
$$\sum E_y=\bf -1930\;\rm N/C$$
The net electric field is given by
$$\sum E=\sqrt{\left(\sum E_x\right)^2+\left(\sum E_y\right)^2}$$
Plug from above,
$$\sum E=\sqrt{\left(9823\right)^2+\left(-1930\right)^2}$$
$$\sum E=\color{red}{\bf 1.00 \times 10^4}\;\rm N/C$$
Its direction is given by
$$\theta_{E}=\tan^{-1}\left[ \dfrac{\sum E_y}{\sum E_x}\right]=\tan^{-1}\left[ \dfrac{-1930}{9823}\right]$$
It is in the fourth quadrant.
$$\theta_{E}=\color{red}{\bf -11.12^\circ}\tag{Below $+x$-direction}$$
clockwise from $+x$-direction.
