Given, the radius of each small drop is $r = 0.17 \times 10^{-2} \text{ m}$. The number of small drops is $n = 25$. The surface tension of water is $T = 7.2 \times 10^{-3} \text{ N/m}$. Let $R$ be the radius of the large drop formed by merging the 25 small drops. Since the total volume remains constant: $$ \begin{aligned} \frac{4}{3} \pi R^3 &= 25 \times \frac{4}{3} \pi r^3 \\ \Rightarrow R^3 &= 25 r^3 \\ \Rightarrow R &= r \sqrt[3]{25} \\ \Rightarrow R &= 0.17 \times 10^{-2} \times 2.924 \\ \Rightarrow R &= 0.497 \times 10^{-2} \text{ m} \end{aligned} $$ The change in surface area $\Delta A$ is the difference between the total surface area of the small drops and the surface area of the large drop: $$ \begin{aligned} \Delta A &= n(4 \pi r^2) - 4 \pi R^2 \\ \Delta A &= 4 \pi (25 r^2 - R^2) \\ \Delta A &= 4 \pi \left[ 25(0.17 \times 10^{-2})^2 - (0.497 \times 10^{-2})^2 \right] \\ \Delta A &= 4 \pi \left[ 7.225 \times 10^{-4} - 2.470 \times 10^{-4} \right] \\ \Delta A &= 4 \pi (4.755 \times 10^{-4}) \\ \Delta A &= 5.974 \times 10^{-4} \text{ m}^2 \end{aligned} $$ The energy released $E$ is given by $E = T \Delta A$: $$ \begin{aligned} E &= 7.2 \times 10^{-3} \times 5.974 \times 10^{-4} \\ E &= 4.301 \times 10^{-6} \text{ J} \end{aligned} $$ Thus, the energy released is approximately $4.3 \times 10^{-6} \text{ J}$.