Answer
$\approx 12\;\rm kN/C$,
$9.3^\circ$
Work Step by Step
The two-point charges are positive, so the electric field direction is away from them.
The upper charge exerting on the dot an electric force which 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 on the dot an electric force which is north east. The angle of this electric field is given by, see the graph below,
$$\tan\theta_2=\dfrac{10}{5}=2\\
\rightarrow \theta_2=\bf 63.4^\circ$$
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,
$$\sum E_x=(8.99\times 10^9)(3\times 10^{-9})\left[ \dfrac{\cos0^\circ}{0.05^2}+\dfrac{\cos63.4^\circ }{(0.05^2+0.1^2)}\right]$$
$$\sum E_x=\bf 11754\;\rm N/C$$
By the same approach, the net electric field in $y$-direction is given by
$$\sum E_y=kq\left[ \dfrac{\sin\theta_1 }{r_1^2}+\dfrac{\sin\theta_2}{r_2^2}\right]$$
$$\sum E_y=(8.99\times 10^9)(3\times 10^{-9})\left[ \dfrac{\sin0^\circ}{0.05^2}+\dfrac{\sin63.4^\circ }{(0.05^2+0.1^2)}\right]$$
$$\sum E_y=\bf 1929\;\rm N/C$$
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,
$$\sum E=\sqrt{\left(11754\right)^2+\left(1929 \right)^2}$$
$$\sum E=\color{red}{\bf 1.19\times 10^4}\;\rm N/C$$
And its direction is given by
$$\theta_{E}=\tan^{-1}\left[ \dfrac{\sum E_y}{\sum E_x}\right]=\tan^{-1}\left[ \dfrac{1929 }{11754}\right]$$
$$\theta_{E}=\color{red}{\bf 9.3^\circ}\tag{Above $+x$-direction}$$
counterclockwise from $+x$-direction.
