Answer
$F = 2.77~N$
Work Step by Step
We can find the y component of the electrostatic force on particle 1 due to particle 2 and particle 3:
$F_y = -\frac{q_1~q_2}{4\pi~\epsilon_0~r^2}-\frac{q_1~q_3}{4\pi~\epsilon_0~r^2}~cos~60^{\circ}$
$F_y = -\frac{(20.0\times 10^{-6}~C)(20.0\times 10^{-6}~C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~(1.50~m)^2}~(1+cos~60^{\circ})$
$F_y = -2.397~N$
We can find the horizontal component of the electrostatic force on particle 1 due to particle 3:
$F_x = \frac{q_1~q_3}{4\pi~\epsilon_0~r^2}~sin~60^{\circ}$
$F_x = \frac{(20.0\times 10^{-6}~C)(20.0\times 10^{-6}~C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~(1.50~m)^2}~sin~60^{\circ}$
$F_x = 1.384~N$
We can find the magnitude of the electrostatic force on particle 1 due to particle 2 and particle 3:
$F = \sqrt{(1.384~N)^2+(-2.397~N)^2}$
$F = 2.77~N$
