Let q be the charge on the bubble, then

$\text{V}=\frac{\text{Kq}}{\text{a}}$  (Here $\text{K}=\frac{1}{4\pi {\epsilon }_0}$ )

$\therefore \text{q}=\frac{\text{Va}}{\text{K}}$

Let after collapsing, the radius of droplet becomes R, then equating the volume, we have

$\left(4\pi {\text{a}}^2\right)\text{t}=\frac{4}{3}\pi {\text{R}}^3$

$\therefore \text{R}={\left(3{\text{a}}^2\text{t}\right)}^{1/3}$

Now, potential of droplet will be

${\text{V}}^{\prime }=\frac{\text{Kq}}{\text{R}}$

Substituting the values, we have

${\text{V}}^{\prime }=\frac{\left(\text{K}\right)\left(\frac{\text{Va}}{\text{K}}\right)}{{\left(3{\text{a}}^2\text{t}\right)}^{1/3}}$

or ${\text{V}}^{\prime }=\text{V}{\left(\frac{\text{a}}{3\text{t}}\right)}^{1/3}$