The workdone in forming a soap film of size $10\mathrm{cm}\times 10\mathrm{cm}$ will be, if the surface tension of soap solution is $3\times {10}^{-2}\mathrm{N}/\mathrm{m}$

The film's surface area is, but since we know that soap films have two surfaces, the film's surface area is,

$\mathrm{\Delta A}=2\times 10\times 10{\mathrm{cm}}^2$

$\mathrm{\Delta A}=2\times {10}^{-2}{\mathrm{cm}}^2$

$\mathrm{Surface} \mathrm{tension}, \mathrm{T}=3\times {10}^{-2}\mathrm{N}/\mathrm{m}$

We know that, the surface tension and surface area product provides the energy needed to build a soap film.

$\mathrm{E}=\mathrm{T}\times \mathrm{\Delta A}$

$\mathrm{E}=3\times {10}^{-2}\times 2\times {10}^{-2}$

$\mathrm{E}=6\times {10}^{-4}\mathrm{J}$

Hence the correct answer is $6\times {10}^{-4}J.$