Let $\vec{r}_1$ and $\vec{r}_2$ are the position vectors for points $P$ and Q then
$\overrightarrow{\mathrm{r}}_1=3 \hat{\mathrm{k}}$ and $\mathrm{r}_1=3 \mathrm{~cm}=0.03 \mathrm{~m}$
$\overrightarrow{\mathrm{r}}_2=4 \hat{\mathrm{j}}$ and $\mathrm{r}_2=4 \mathrm{~cm}=0.04 \mathrm{~m}$
Position of point $\mathrm{R}$ does not affect the result.
Also $\mathrm{q}=8 \mathrm{mC}=8 \times 10^{-3} \mathrm{C}$.
By formula, $\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}}$
For point $\mathrm{P}$,
$$
\mathrm{V}_1=\frac{9 \times 10^9 \times 8 \times 10^{-3}}{0.03}=24 \times 10^8 \mathrm{~V}
$$
For point $Q$,
$$
\mathrm{V}_2=\frac{9 \times 10^9 \times 8 \times 10^{-3}}{0.04}=18 \times 10^8 \mathrm{~V}
$$
Potential difference, $\left(\mathrm{V}_2-\mathrm{V}_1\right)$
$$
=18 \times 10^8-24 \times 10^8=-6 \times 10^8 \mathrm{~V}
$$
Work done $\mathrm{W}=$ charged moved $\times$ potential difference, $=\left(-2 \times 10^{-9}\right)\left(-6 \times 10^8\right)=1.2 \mathrm{~J}$