A liquid drop of diameter ‘D’ is split into 27 droplets. If T is surface tension of the liquid the change in energy is 

From the question volume is conserved,

Thus the initial volume is $\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^2$

final volume is $\frac{4}{3}{\mathrm{\pi r}}^{2}27$

By volume of conservation,

$\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^2=\frac{4}{3}{\mathrm{\pi r}}^2\Rightarrow \mathrm{r}=\frac{\mathrm{D}}{6}$

Thus the initial energy = $4\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^2\mathrm{T}$

final energy $=4\mathrm{\pi }{\left(\frac{\mathrm{D}}{6}\right)}^2\mathrm{T}\times 27$

Energy change = final energy - initial energy.

$=4\mathrm{\pi }{\left(\frac{\mathrm{D}}{6}\right)}^2\times \mathrm{T}\times 27-4\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^2\times \mathrm{T}$

$=4{\mathrm{\pi TD}}^2\left[\frac{27}{36}-\frac{1}{4}\right]=4{\mathrm{\pi TD}}^2\left[\frac{18}{36}\right]=2{\mathrm{\pi TD}}^2$

Hence the correct answer is $2{\mathrm{\pi TD}}^2.$