Answer
$\overrightarrow{F_{3 \space net}} = (89.9 N) \hat{i}$
Work Step by Step
We have the following information
$q_1 = -80.0 \mu C$
$q_2 = + 40.0 \mu C $
$q_3 = 20 \mu C$
Now we will describe all the charges in vector form with the purpose of finding what is the net force on the third charge, $q_{3 \space net}$
$\overrightarrow{F_{3 \space net}} = \overrightarrow{F_{31}} + \overrightarrow{F_{32}} $
Where
$\overrightarrow{F_{31}} = \frac{kq_3|q_1|}{(r_{31})^2}$ and $\overrightarrow{F_{32}} = \frac{kq_3|q_2|}{(r_{32})^2}$
$ r_{32} = 0.2 m $ directed towards +x
$ r_{31} = 0.4 m $ directed towards -x
So,
$\overrightarrow{F_{3 \space net}} = -|\overrightarrow{F_{31}}| + |\overrightarrow{F_{32}}| $
$\overrightarrow{F_{3 \space net}} = - \frac{kq_3|q_1|}{(r_{31})^2}\hat{i} + \frac{kq_3q_2}{(r_{32})^2} \hat{i}$
$\overrightarrow{F_{3 \space net}} = ( - \frac{kq_3|q_1|}{(r_{31})^2} + \frac{kq_3q_2}{(r_{32})^2}) \hat{i}$
$\overrightarrow{F_{3 \space net}} = kq_3 (-\frac{|q_1|}{(r_{31})^2} + \frac{q_2}{(r_{32})^2} ) \hat{i}$
$\overrightarrow{F_{3 \space net}} = (8.99 \times 10^9 N . m^2 / C^2) (20 \mu C) (-\frac{|-80.0 \mu C|}{(0.4 m)^2} + \frac{40.0 \mu C}{( 0.2 m )^2} ) \hat{i}$
$\overrightarrow{F_{3 \space net}} = ( 179800 \frac{N.m^2}{C})(0.0005 \frac{C}{m^2})\hat{i} $
$\overrightarrow{F_{3 \space net}} = (89.9 N) \hat{i}$
