Solution:

Charge located at the origin, $q=8 \mathrm{mC}=8 \times 10^{-3} \mathrm{C}$

The magnitude of the charge transferred from the point $P$ to the point $R$ 
$Q, q_{1}=-2 \times 10^{-9} \mathrm{C}$ Here $O P=d_{1}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$

$Q=d_{2}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$

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

Potential at the point $\mathrm{Q}, V_{2}=\frac{q}{4 \pi \epsilon \theta d_{2}}$

The work done $(W)$ is independent of the path

Therefore, $W=q_{1}\left[V_{1}-V_{2}\right]$

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

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

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

Therefore,

$W=9 \times 10^{9} \times 8 \times 10^{-3} \times\left(-2 \times 10^{-9}\right)\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{3 \times 10^{-2}}\right]$

$=-144 \times 10^{-3} \times(-100 / 12)$ $=1.2 \mathrm{~J}$ oule

Therefore, the work done during the process is $1.2 \mathrm{~J}$