Answer
The force on the 1.0 nC charge is $1.8\times 10^{-4}~N$ directed to the right.
Work Step by Step
By symmetry, the vertical component of each force on the 1.0 nC charge cancels out. The net force on the 1.0 nC charge is the sum of the horizontal component of each force. The horizontal component of each force on the 1.0 nC charge is directed to the right.
$F = 2\times \frac{(k)(1.0~nC)(2.0~nC)}{r^2}~cos(60^{\circ})$
$F = 2\times \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{(0.010~m)^2}~cos(60^{\circ})$
$F = 1.8\times 10^{-4}~N$
The force on the 1.0 nC charge is $1.8\times 10^{-4}~N$ directed to the right.
