Let radius of the newly formed drops be $r$.

As total volume will remain constant, therefore

$\frac{4}{3}\pi R^3=27\times \left(\frac{4}{3}\times \pi r^3\right)$

$\Rightarrow R^3=27r^3$

$\Rightarrow R=3r$

$\Rightarrow r=\frac{R}{3}$

$\Rightarrow r^2=\frac{R^2}{9}$

Now, work done $=T\times \Delta A$

$$\begin{gathered} =T\left[27 \left(4\pi r^2\right)\right]-T\left[4\pi R^2\right] \\ =T\left[27 \left(4\pi \frac{R^2}{9}\right)\right]-T\left[4\pi R^2\right] \\ =T\pi R^2\left(12-4\right) \end{gathered}$$

$=8\pi R^{2}T$