Sixty-four drops of water, each of radius 0.50 mm, join to make one big drop…

Step 1. Radius. $64^{1/3} = 4$, so $R = 2.0\ \text{mm}$.

Step 2. Area lost. $(64 - 16) \times 4\pi(5.0 \times 10^{-4}\ \text{m})^2 = 1.51 \times 10^{-4}\ \text{m}^2$.

Step 3. Energy. $0.072\ \text{N/m} \times 1.51 \times 10^{-4}\ \text{m}^2 = 1.09 \times 10^{-5}\ \text{J}$.