Excess pressure inside a soap bubble of radius R is

$=\frac{4T}{R}$     .....(i)

Where T is surface tension of liquid film.

Pressure due to oil column

$=h\mathrm{\rho }\mathrm{g}$      .....(ii)

Where $h$ is height of column, $\mathrm{\rho }$ the density and $g$ the gravity.

From Eqs. (i) and (ii), we get

$\frac{4T}{R}=h\mathrm{\rho }\mathrm{g}$

$⟹ T=\frac{h\mathrm{\rho }\mathrm{g}R}{4}$

$\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n},\mathrm{ }\mathrm{ }\mathrm{ }h=2 \mathrm{m}\mathrm{m,}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{g}=980 \mathrm{c}\mathrm{m} {\mathrm{s}}^{-2},$

$\mathrm{\rho }=0.8\mathrm{ }\mathrm{g}\mathrm{c}{\mathrm{c}}^{-1},\mathrm{ }R=1 \mathrm{c}\mathrm{m}$

$\therefore T=\frac{0.2\times 0.8\times 980}{4}$

$=\frac{2\times {10}^{-3}\times 0.8\times {10}^3\times 10\times 1\times {10}^{-2}}{4} \mathrm{N} {\mathrm{m}}^{-1}$

$=4\times {10}^{-2} \mathrm{N} {\mathrm{m}}^{-1}$