Answer
$F_{3}=9.96 \times 10^{-12} \mathrm{~N} .
$
Work Step by Step
The net force component along the $x$ axis points rightward. With $\theta=60^{\circ}$,
$
F_3=2 \frac{q_3 q_1 \cos \theta}{4 \pi \varepsilon_0 a^2} .
$
Since $\cos \left(60^{\circ}\right)=1 / 2$, we can write this as
$
F_3=\frac{k q_3 q_1}{a^2}=\frac{\left(8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{C}^2\right)\left(5.00 \times 10^{-12} \mathrm{C}\right)\left(2.00 \times 10^{-12} \mathrm{C}\right)}{(0.0950 \mathrm{~m})^2}\\=9.96 \times 10^{-12} \mathrm{~N} .
$
