#### Answer

The net force on the +1.0 nC charge is $1.2\times 10^{-5}~N$ directed upward.

#### Work Step by Step

By symmetry, the horizontal 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 vertical component of each force.
We can find the magnitude of the upward force on the +1.0 nC charge by the -6.0 nC charge.
$F_u = \frac{(k)(1.0~nC)(6.0~nC)}{r^2}$
$F_u = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(6.0\times 10^{-9}~C)}{(0.050~m)^2}$
$F_u = 2.2\times 10^{-5}~N$
We can find the downward force on the +1.0 nC charge which is the sum of the vertical component from each of the two +2.0 nC charges.
$F_d = 2\times \frac{(k)(1.0~nC)(2.0~nC)}{r^2}~cos(45^{\circ})$
$F_d = 2\times \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{(0.050~m)^2}~cos(45^{\circ})$
$F_d = 1.0\times 10^{-5}~N$
We can find the net upward force on the +1.0 nC charge.
$F = F_u-F_d$
$F = 2.2\times 10^{-5}~N - 1.0\times 10^{-5}~N$
$F = 1.2\times 10^{-5}~N$
The net force on the +1.0 nC charge is $1.2\times 10^{-5}~N$ directed upward.