If work done in increasing the size of soap film from $10 \mathrm{cm}\times 6 \mathrm{cm}$ to $10 \mathrm{cm}\times 11\mathrm{cm}$ is $2\times {10}^{-4}\mathrm{J}$, then the surface tension is

Both of the film's faces are affected by surface tension. Therefore, the change in area of both surfaces must be taken into account.

$\mathrm{Work} \mathrm{done} = \mathrm{Area} \mathrm{increased} \times \mathrm{surface} \mathrm{tension}$

${\mathrm{S}}_{\mathrm{T}}=\frac{2\times {10}^{-4}}{2\times (10\times 11-10\times 6)\times {10}^{-4}}=2\times {10}^{-2}{\mathrm{Nm}}^{-1}$

Hence the correct answer is $2\times {10}^{-2}{\mathrm{Nm}}^{-1}.$