Let $R$ be radius of big drop
Then $\large\frac{4}{3} $$\pi R^3=1000\large\frac{4}{3}$$ \times \pi \times (10^{-7})^3$
$R= 10^{-6}$
Changes in surface area
$\Delta A= 4 \pi R^2- 1000 \times 4 \pi r^2$
$\qquad= 4 \pi \bigg[(10^{-6})^2 -10^3 \times (10^{-7})^2\bigg]$
$\qquad =-36 \pi (10^{-12}) m^2$
Energy released $=T \times |\Delta A|$
$\qquad=14 \times 10^{-2} \times 36 \times \large\frac{22}{7}$$ \times 10^{-12}$
$\qquad=1.58 \times 10^{-11}\;J$
Hence b is the correct answer.