Magnitude of a small charge, which is taken from a point $P$ to point $R$ to point $Q, \;\;q_{1}=-2 \times 10^{-9} \,C$

All the points are represented in the given figure.

Point $P$ is at a distance, $d_{1}=3 \,cm ,$ from the origin along $z$ -axis. Point $Q$ is at a distance, $d _{2}=4 \,cm ,$ from the origin along $y$ -axis.

Potential at point $P, \quad V_{1}=\frac{q}{4 \pi \epsilon_{0} \times d_{1}}$

Potential at point $Q$, $\quad V_{2}=\frac{q}{4 \pi \epsilon_{0} d_{2}}$

Work done $(W)$ by the electrostatic force is independent of the path. $\therefore W=q_{1}\left[V_{2}-V_{1}\right]$

$=q_{1}\left[\frac{q}{4 \pi \epsilon_{0} d_{2}}-\frac{q}{4 \pi \epsilon_{0} d_{1}}\right]$

$=\frac{q q_{1}}{4 \pi \epsilon_{0}}\left[\frac{1}{d_{2}}-\frac{1}{d_{1}}\right]$

Where, $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \,N\, m ^{2} \,C ^{-2}$

$\therefore W=9 \times 10^{9} \times 8 \times 10^{-3} \times\left(-2 \times 10^{-9}\right)\left[\frac{1}{0.04}-\frac{1}{0.03}\right]$

$=-144 \times 10^{-3} \times\left(\frac{-25}{3}\right)$

$=1.27 \,J$

Therefore, work done during the process is $1.27 \;J$