Answer
$F=1.152\times10^{-23}\;N$
Work Step by Step
The net electrostatic force on $q_7$ along positive x-axis is
$F_{+x}=\frac{1}{4\pi\epsilon_0}[\frac{12e^2}{d^2}-\frac{e^2}{d^2}-\frac{24e^2}{4d^2}]$
or, $F_{+x}=\frac{1}{4\pi\epsilon_0}\frac{5e^2}{d^2}$
The net electrostatic force on $q_7$ along positive y-axis is
$F_{+y}=\frac{1}{4\pi\epsilon_0}[\frac{12e^2}{d^2}+\frac{48e^2}{4d^2}-\frac{24e^2}{d^2}]$
or, $F_{+y}=0$
Therefore, the magnitude of the net electrostatic force on particle 7 is
$F=F_{+x}=\frac{1}{4\pi\epsilon_0}\frac{5e^2}{d^2}$
or, $F=\frac{1}{4\pi\epsilon_0}\frac{5\times (1.6\times10^{-19})^2}{0.01^2}\;N$
or, $F=1.152\times10^{-23}\;N$
