Answer
$F = 1.1\times 10^{-5}~N$ (upward)
Work Step by Step
By symmetry the horizontal components of the three forces cancel out.
We can find the upward force on the charge at the bottom:
$F = \frac{k~(6.0~nC) (1.0~nC)}{r^2}- 2\times \frac{k~(2.0~nC)(1.0~nC)}{r^2}~cos~45^{\circ}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(6.0\times 10^{-9}~C)(1.0\times 10^{-9}~C)}{(0.050~m)^2}-2\times \frac{(9.0\times 10^9~N~m^2/C^2)(2.0\times 10^{-9}~C)(1.0\times 10^{-9}~C)}{(0.050~m)^2}~cos~45^{\circ}$
$F = 1.1\times 10^{-5}~N$
