Answer
$F = 1.60~N$
Work Step by Step
We can find the magnitude of the electrostatic force on particle 1 due to particle 2:
$F = \frac{q_1~q_2}{4\pi~\epsilon_0~r^2}$
$F = \frac{(20.0\times 10^{-6}~C)(20.0\times 10^{-6}~C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~(1.50~m)^2}$
$F = 1.60~N$
