The work done will be against surface tension and we know that surface tension is related to work as $\mathrm{W}=\mathrm{T}\times ∆\mathrm{A}$
Since, the surface area of sphere is $4{\mathrm{\pi R}}^2$ and there are two free surfaces, we have

$\mathrm{W}=\mathrm{T}\times 8{\mathrm{\pi R}}^2$      ...$\left(\mathrm{i}\right)$
and volume of sphere $=\frac{4}{3}{\mathrm{\pi R}}^3$

i.e. $\mathrm{V}=\frac{4}{3}{\mathrm{\pi R}}^3$

$\Rightarrow \mathrm{R}={\left(\frac{3\mathrm{V}}{4\mathrm{\pi }}\right)}^{1/3}$   ...$\left(\mathrm{ii}\right)$

From Eqs. $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get

$\mathrm{W}=\mathrm{T}\times 8\mathrm{\pi }\times {\left(\frac{3\mathrm{V}}{4\mathrm{\pi }}\right)}^{2/3}$

$\Rightarrow \mathrm{ }\mathrm{W}\propto {\mathrm{V}}^{2/3}$

$\therefore {\mathrm{W}}_1\propto {\mathrm{V}}_1^{2/3}$

and, ${\mathrm{W}}_2\propto {\mathrm{V}}_2^{2/3}$

$\frac{{\mathrm{W}}_2}{{\mathrm{W}}_1}={\left(\frac{2{\mathrm{V}}_1}{{\mathrm{V}}_1}\right)}^{2/3}$

$\Rightarrow {\mathrm{W}}_2=2^{2/3}{\mathrm{W}}_1=4^{1/3}\mathrm{W}$