D

Step 1: Understand the process of coalescence of water drops. When two water drops merge to form a larger drop, the surface tension plays a crucial role in this process.

Step 2: Analyze energy changes. During coalescence, the total surface energy of the system changes because smaller drops have a higher surface area to volume ratio compared to a larger drop. The energy associated with the surface tensions of the drops may be released when they coalesce into a larger drop.

Step 3: Evaluate surface areas. If we denote the radius of the smaller drops as \( r_1 \) and \( r_2 \), then the total surface area of the two smaller drops is given by:
\( A_1 = 4\pi r_1^2 \) and \( A_2 = 4\pi r_2^2 \).
The radius \( R \) of the resulting drop after coalescence can be derived using volume conservation:
\( V_{total} = V_1 + V_2 \Rightarrow (\frac{4}{3}\pi R^3 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3) \Rightarrow R^3 = r_1^3 + r_2^3 \).

Step 4: The surface area of the bigger drop is:
\( A = 4\pi R^2 \). It can be shown that for two coalescing drops, \( R < r_1 + r_2 \), meaning the surface area of the larger drop is less than the sum of the surface areas of the two smaller drops.

Therefore, the correct answer is that the surface area of the bigger drop is smaller than the sum of the surface areas of both drops.
Thus, the correct answer is option D.