Answer
$\overrightarrow{F_{3 \space net}} = (0.829N) \hat{i}$
Work Step by Step
$Q_1 = 80.0 nC$ at xy coordinates $(0, 0.003 m)$
$Q_2 = 80.0 nC$ at xy coordinates $ (0, -0.003 m) $
$Q_3 = 18.0 nC$ at xy coordinates $ (0.004 m, 0)$
Now we want to find out what is the $\overrightarrow{F_{3 \space net}} $
$\overrightarrow{F_{3 \space net}} = \overrightarrow{F_{31}} + \overrightarrow{F_{32}} $
Where
$\overrightarrow{F_{31}} = \frac{kq_3|q_1|}{(r_{31})^2}$ and $\overrightarrow{F_{32}} = \frac{kq_3|q_2|}{(r_{32})^2}$
From Pythagoras Theorem,
$ r_{31} = \sqrt {(0.003)^2 +(0.004)^2}$
$ r_{31} = 0.005 m$
$ r_{31} = r_{32} $ because the magnitudes are the same.
$\overrightarrow{F_{3 \space net}} = - \frac{kq_3|q_1|}{(r_{31})^2}\hat{i} + \frac{kq_3q_2}{(r_{32})^2} \hat{i}$
In magnitude-angle notation (particularly convenient if one uses a vector capable calculator in polar mode), the indicated vector addition becomes
$\overrightarrow{F_{3 \space net}} = (0.518 \angle - 37^o) +( 0.518\angle37^o) = 0.829\angle0^o $
$\overrightarrow{F_{3 \space net}} = (0.829N) \hat{i}$
