#### Answer

The force on the +1.0 nC charge is $1.0\times 10^{-3}~N$ directed to the left.

#### Work Step by Step

We can find the distance $r$ from the +1.0 nC charge to each of the other charges.
$r = \sqrt{(0.5~cm)^2+(0.5~cm)^2}$
$r = 0.71~cm$
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 left.
$F = 4\times \frac{(k)(1.0~nC)(2.0~nC)}{r^2}~cos(45^{\circ})$
$F = 4\times \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{(0.0071~m)^2}~cos(45^{\circ})$
$F = 1.0\times 10^{-3}~N$
The force on the +1.0 nC charge is $1.0\times 10^{-3}~N$ directed to the left.