Answer
$F = (1.02\times 10^{-5}~N)~\hat{i}+(2.16\times 10^{-5}~N)~\hat{j}$
Work Step by Step
By symmetry, the vertical components of the forces due to the $2.0~nC$ charge and the $-2.0~nC$ charge cancel out.
We can find the upward force on the charge at the bottom:
$F_y = \frac{k~(6.0~nC) (1.0~nC)}{r^2}$
$F_y = \frac{(9.0\times 10^9~N~m^2/C^2)(6.0\times 10^{-9}~C)(1.0\times 10^{-9}~C)}{(0.050~m)^2}$
$F_y = 2.16\times 10^{-5}~N$
We can find the rightward force on the charge at the bottom:
$F_x = 2\times \frac{k~(2.0~nC)(1.0~nC)}{r^2}~cos~45^{\circ}$
$F_x = 2 \times \frac{(9.0\times 10^9~N~m^2/C^2)(2.0\times 10^{-9}~C)(1.0\times 10^{-9}~C)}{(0.050~m)^2}~cos~45^{\circ}$
$F_x = 1.02\times 10^{-5}~N$
We can express the net force in component form:
$F = (1.02\times 10^{-5}~N)~\hat{i}+(2.16\times 10^{-5}~N)~\hat{j}$
