A soap bubble of radius 1.5 cm is blown with a soap solution of surface tension 27 mN/m…

Step 1. The excess pressure. $\Delta P = \dfrac{4\gamma}{r} = \dfrac{4 \times 0.027}{0.015} = 7.2\ \text{Pa}$.

Step 2. The area. One surface has $4\pi r^2 = 4\pi \times (0.015)^2 = 2.827 \times 10^{-3}\ \text{m}^2$. The two surfaces together have $5.65 \times 10^{-3}\ \text{m}^2$.

Step 3. The work. $W = \gamma\,\Delta A = 0.027 \times 5.65 \times 10^{-3} = 1.53 \times 10^{-4}\ \text{J}$, which is $0.000153\ \text{J}$.