The gauge pressure inside a soap bubble of surface tension T is...

Excess pressure inside a soap bubble is given by $\Delta P = \dfrac{4T}{R}$, where $R$ is the radius of the bubble.

Gauge pressure exerted by a liquid column of height $h$ and density $\rho$ is $P = \rho g h$.

Equating the two pressures as given:
$\dfrac{4T}{R} = \rho g h \implies R = \dfrac{4T}{\rho g h}$

The work done in blowing a soap bubble from zero radius is equal to its surface energy. Since a soap bubble has two free surfaces (inner and outer), the work done is:
$W = T \times 2 \times 4\pi R^2 = 8\pi R^2 T$

Substituting the value of $R$ into the work equation:
$W = 8\pi T \left(\dfrac{4T}{\rho g h}\right)^2$

$W = 8\pi T \left(\dfrac{16T^2}{\rho^2 g^2 h^2}\right) = \dfrac{128\pi T^3}{\rho^2 g^2 h^2}$

Note: Using $W = 4\pi R^2 T$ (forgetting the two surfaces for work) leads to the incorrect option $\dfrac{64\pi T^3}{\rho^2 g^2 h^2}$, and using $\Delta P = \dfrac{2T}{R}$ (forgetting two surfaces for pressure) leads to the incorrect option $\dfrac{32\pi T^3}{\rho^2 g^2 h^2}$.

Answer: $\dfrac{128\pi T^3}{\rho^2 g^2 h^2}$