Step 1: Surface energy
The surface energy of a drop is $T\times$ area. The initial energy is $4\pi R^2T$.
Step 2: Final energy
There are $n = (R/r)^3$ droplets, each with energy $4\pi r^2T$, so the final energy is $4\pi r^2T\cdot\dfrac{R^3}{r^3} = \dfrac{4\pi R^3T}{r}$.
Step 3: Difference
Energy needed $= \dfrac{4\pi R^3T}{r} - 4\pi R^2T = 4\pi R^2T\left(\dfrac{R}{r} - 1\right)$.
Step 4: Check
When $r = R$ there is just one drop and the energy needed is zero, which matches the formula.
Final Answer:
The energy is 4 pi T R^2 (R/r - 1). This is option (A).
\[ \boxed{\text{(A) }4\pi TR^2\left[\frac{R}{r}-1\right]} \]