By conserving volume, the volume of the large drop equals the sum of the volumes of 1000 smaller drops: \[\frac{4}{3}\pi R^3 = 1000 \cdot \frac{4}{3}\pi r^3,\] where \( R \) is the radius of the large drop and \( r \) is the radius of each small drop. Simplifying yields: \[\R^3 = 1000 \cdot r^3 \implies R = 10r.\] Surface energy is proportional to surface area. The initial energy (1000 small drops) is: \[\E_i = 1000 \cdot 4\pi r^2.\] The final energy (large drop) is: \[\E_f = 4\pi R^2 = 4\pi (10r)^2 = 100 \cdot 4\pi r^2.\] The ratio of final to initial energy is: \[\frac{E_f}{E_i} = \frac{4\pi (10r)^2}{1000 \cdot 4\pi r^2} = \frac{1}{10}.\] Final Answer: \[\boxed{\frac{1}{10} \, \text{th}}.\]