Two charges + Q and -Q are arranged as shown in fig. The work done in carrying a test charge q from X to Y will be ($1/4{\mathrm{\pi \epsilon }}_0$) times

The potential at Y is given by
${\mathrm{V}}_{\mathrm{Y}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{Q}}{(\mathrm{a}+\mathrm{r})}+\frac{-\mathrm{Q}}{\mathrm{a}}\right]$
The potential at X is given by
${\mathrm{V}}_{\mathrm{X}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{Q}}{\mathrm{a}}+\frac{-\mathrm{Q}}{(\mathrm{a}+\mathrm{r})}\right]$
$$\begin{gathered} \text{ Now }\mathrm{W}=\mathrm{q}\left[{\mathrm{V}}_{\mathrm{Y}}-{\mathrm{V}}_{\mathrm{X}}\right] \\ =\frac{2\mathrm{qQ}}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{(\mathrm{a}+\mathrm{r})}-\frac{1}{\mathrm{a}}\right] \\ \therefore \left|\mathrm{W}\right|=\frac{2\mathrm{qQr}}{\mathrm{a}(\mathrm{a}+\mathrm{r})}\times \frac{1}{4{\mathrm{\pi \epsilon }}_0} \end{gathered}$$