An air bubble of radius R is formed at a depth h in a liquid of...

The absolute pressure inside an air bubble of radius $R$ at a depth $h$ in a liquid is the sum of atmospheric pressure, hydrostatic pressure, and the excess pressure (which has only one free surface):
$P_{\text{air bubble}} = P_0 + \rho g h + \dfrac{2T}{R}$

The absolute pressure inside a soap bubble of radius $2R$ in the atmosphere (which has two free surfaces) is:
$P_{\text{soap bubble}} = P_0 + \dfrac{4T'}{(2R)} = P_0 + \dfrac{2T'}{R}$

According to the problem, these two absolute pressures are equal:
$P_0 + \rho g h + \dfrac{2T}{R} = P_0 + \dfrac{2T'}{R}$

Cancelling $P_0$ from both sides:
$\rho g h + \dfrac{2T}{R} = \dfrac{2T'}{R}$

Multiplying the entire equation by $\dfrac{R}{2}$ to isolate $T'$:
$T' = \dfrac{\rho g h R}{2} + T$

Answer: $T + \dfrac{\rho g h R}{2}$