Work of $3.0\times {10}^{-4}$ joule is required to be done in increasing the size of a soap film from $10\mathrm{cm}\times 6\mathrm{cm} \mathrm{to} 10\mathrm{cm}\times 11\mathrm{cm}$. The surface tension of the film is

The surface tension is 

$\mathrm{S}=\frac{\mathrm{work} \mathrm{done}}{\mathrm{increase} \mathrm{in} \mathrm{area}}$

from the question as we know that area increases by

$2\left[\left(11\times 10\right)-\left(10\times 6\right)\right]\times {10}^{-4} {\mathrm{m}}^2$

As soap film has two surfaces

$\mathrm{S}=\frac{3\times {10}^{-4}}{2\times 50\times {10}^{-4}}$

$\mathrm{S}=3\times {10}^{-2} \mathrm{N}/\mathrm{m}$