The volume of the bigger drop will be, V=34πR3. If the radius of smaller drops is r, using the conservation of volume,
n34πr3=34πR3⇒729×34πr3=34πR3⇒r=9R
The initial surface energy will be,
Ei=(4πR2)T
And final surface energy will be,
Ef=n(4πr2)T=729×4π(9R)2T=36πR2T
Therefore, the change in the surface energy will be,
⇒ΔE=Ef−Ei=32πR2T=32×3.14×(1×10−2)2×75×10−3⇒ΔE=7.5×10−4J