Answer
$4.3\times 10^{-3}\;\rm N$, $252.4^\circ$
Work Step by Step
First of all, we need to draw the force diagram exerted on the -10 nC-charge. Recalling that the like charges repel and the unlike charge attract.
As seen below, we put the -10 nC-charge at the origin and we numbered the three charges from 1 to 3.
The net electric force exerted on the charge $q_1=\rm-10 nC$ in the $x$-direction is given by
$$\sum F_x=F_{x,2\rm on1}-F_{x,3\rm on1}$$
We will neglect the signs of the charges since we are focusing on the charge's force direction.
$$\sum F_x=\dfrac{kq_1q_2}{r^2_{2\rightarrow1}}\cos(-90^\circ)-\dfrac{kq_1q_3}{r^2_{3\rightarrow1}}\cos\theta$$
where $\cos\theta=\dfrac{x_3}{r_{3\rightarrow1}}$ and $r=\sqrt{x_3^2+y_3^2}$, so that
$$\sum F_x=0-\dfrac{kq_1q_3x_3}{(x_3^2+y_3^2)^\frac{3}{2}}$$
Plug the known;
$$\sum F_x=0-\dfrac{(8.99\times 10^9)(10\times 10^{-9})(15\times 10^{-9})(0.03)}{(0.03^2+0.01^2)^\frac{3}{2}}$$
$$\sum F_x=\bf -1.3\times10^{-3}\;\rm N$$
The net electric force exerted on the charge $q_1=\rm-10 nC$ in the $y$-direction is given by
$$\sum F_y=-F_{y,2\rm on1}+F_{y,3\rm on1}$$
$$\sum F_y=-\dfrac{kq_1q_2}{r^2_{2\rightarrow1}}\sin(90^\circ)+\dfrac{kq_1q_3}{r^2_{3\rightarrow1}}\sin\theta$$
where $\sin\theta=\dfrac{y_3}{r_{3\rightarrow1}}$ and $r=\sqrt{x_3^2+y_3^2}$, so that
$$\sum F_y=-\dfrac{kq_1q_2}{r^2_{2\rightarrow1}} +\dfrac{kq_1q_3y_3}{r^3_{3\rightarrow1}}$$
$$\sum F_y=-\dfrac{(8.99\times 10^9)(10\times 10^{-9})(5\times 10^{-9})}{0.01^2} +\dfrac{(8.99\times 10^9)(10\times 10^{-9})(15\times 10^{-9})(0.01)}{{(0.03^2+0.01^2)^\frac{3}{2}} }$$
$$\sum F_y=\bf -4.1\times 10^{-3}\;\rm N$$
Hence, the magnitude of the net force is given by
$$\sum F_{\rm on 1}=\sqrt{(\sum F_x)^2+(\sum F_y)^2}$$
$$\sum F_{\rm on 1}=\sqrt{(-1.3\times 10^{-3})^2+(-4.1\times 10^{-3})^2}$$
$$\sum F_{\rm on 1}=\color{red}{\bf 4.3\times 10^{-3}}\;\rm N$$
And its direction is given by
$$\tan\theta_{net}=\dfrac{\sum F_y}{\sum F_x}$$
where $\theta_{net}$ is in the third quadranet.
$$\theta_{net}=\tan^{-1}\left[\dfrac{\sum F_y}{\sum F_x}\right]=\tan^{-1}\left[\dfrac{-4.1\times 10^{-3}}{-1.3\times 10^{-3}}\right]=72.4^\circ$$
And then its direction from $+x$-direction is
$$\theta_{net}=180^\circ +72.4^\circ$$
$$\theta_{net}=\color{red}{\bf 252.4^\circ}\tag{CCW}$$
Counterclockwise from $+x$-direction.
