#### Answer

The net force on charge A is zero.

#### Work Step by Step

We can find the magnitude of the electric force on A from B:
$F = \frac{k~q_A~q_B}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(1.0\times 10^{-9}~C)}{(0.010~m)^2}$
$F = 9.0\times 10^{-5}~N$
A force of $9.0\times 10^{-5}~N$ is directed to the right.
We can find the magnitude of the electric force on A from C:
$F = \frac{k~q_A~q_C}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(4.0\times 10^{-9}~C)}{(0.020~m)^2}$
$F = 9.0\times 10^{-5}~N$
A force of $9.0\times 10^{-5}~N$ is directed to the left.
Since the same magnitude of force is directed to the right and left, the net force on charge A is zero.