If the work done in blowing a soap bubble of volume 'v' is w, then the work done in blowing a soap bubble of volume '2v' is

We know that work done is,

$\mathrm{W}=\mathrm{T}△\mathrm{A}, \mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3$

$\mathrm{Area}=4{\mathrm{\pi r}}^2$

$\Rightarrow \mathrm{A}\propto {\mathrm{v}}^{\frac{2}{3}}$

Thus,

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

${\mathrm{W}}_2={\mathrm{W}}_1{\left(\frac{2\mathrm{V}}{\mathrm{V}}\right)}^{\frac{2}{2}}\Rightarrow {\mathrm{W}}_2{2}^{\frac{2}{3}}=4^{\frac{1}{3}}\mathrm{W}$

Hence the correct answer is $4^{\frac{1}{3}}\mathrm{W}.$