Answer
The maximum magnitude is $~~F = 4.90\times 10^{-26}~N$
Work Step by Step
When $0 \leq x \leq 5.0~m$, the vertical components of the forces cancel out.
When $0 \lt x \leq 5.0~m$, the horizontal components of the forces are directed in the positive direction of the x axis. The net force on particle 3 is the sum of the horizontal components.
In part (b), we found that there is a maximum at $~~x = 12.0~cm$
We can find $F$ when $x = 12.0~cm$:
$F = 2\times \frac{q_1~q_3}{4\pi~\epsilon_0~(\sqrt{x^2+d^2})^2}~cos~\theta$
$F = \frac{q_1~q_3}{2\pi~\epsilon_0~(x^2+d^2)}~\frac{x}{\sqrt{x^2+d^2}}$
$F = \frac{q_1~q_3}{2\pi~\epsilon_0}~\frac{x}{(x^2+d^2)^{3/2}}$
$F = \frac{(3.20\times 10^{-19}~C)(6.40\times 10^{-19}~C)}{(2\pi)~(8.854\times 10^{-12}~F/m)}~\frac{0.120~m}{[(0.120~m)^2+(0.170~m)^2]^{3/2}}$
$F = 4.90\times 10^{-26}~N$
The maximum magnitude is $~~F = 4.90\times 10^{-26}~N$
