Answer
The magnitude of the net electrostatic force on the $CL^{-}$ ion is $~~1.9\times 10^{-9}~N$
Work Step by Step
By symmetry, the electrostatic forces on the $CL^{-}$ ion due to two $Cs^{+}$ ions in opposite corners of the cube cancel out.
For six of the $Cs^{+}$ ions, there is another $Cs^{+}$ ion in the opposite corner.
Therefore, the net electrostatic force on the $CL^{-}$ ion due to these six $Cs^{+}$ ions is zero.
However, there is one $Cs^{+}$ ion without a $Cs^{+}$ ion in the opposite corner.
The magnitude of the net electrostatic force on the $CL^{-}$ ion is equal to the magnitude of the electrostatic force on the $CL^{-}$ ion due to this one $Cs^{+}$ ion.
We can find the magnitude of the force:
$F = \frac{\vert q_1 \vert~\vert q_2 \vert}{4\pi~\epsilon_0~r^2}$
$F = \frac{e^2}{4\pi~\epsilon_0~r^2}$
$F = \frac{(1.6\times 10^{-19}~C)(1.6\times 10^{-19}~C)}{(4\pi)~(8.854\times 10^{-12}~F/m)~[(\sqrt{3})(0.20)\times 10^{-9}~m]^2}$
$F = 1.9\times 10^{-9}~N$
The magnitude of the net electrostatic force on the $CL^{-}$ ion is $~~1.9\times 10^{-9}~N$
