The amount of work done to break a big water drop of radius  $$' R '$$  into  27  small drops of equal radius is  $$10\,J.$$ The work done required to break the same big drop into  64  small drops of equal radius will be:

Work to break a drop of radius R into 27 drops is 10 J and we need the work to break it into 64 drops.

Since the work is given by $$W = T \times \Delta A = T(n \times 4\pi r^2 - 4\pi R^2),$$

and volume conservation implies $$n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow r = R/n^{1/3},$$

substituting in yields $$W = 4\pi TR^2(n^{1/3}-1).$$

For 27 drops $$W_1 = 4\pi TR^2(27^{1/3}-1) = 4\pi TR^2(3-1) = 8\pi TR^2 = 10$$ J

For 64 drops $$W_2 = 4\pi TR^2(64^{1/3}-1) = 4\pi TR^2(4-1) = 12\pi TR^2$$

This gives $$\frac{W_2}{W_1} = \frac{12\pi TR^2}{8\pi TR^2} = \frac{3}{2},$$

therefore $$W_2 = \frac{3}{2} \times 10 = 15$$ J