When a big drop of radius $R$, breaks into $N$ droplets each of radius $r$, the volume remains constant. So by the law of conservation of mass, final volume $=$ initial volume. Volume of big drop $=N \times$ Volume of each small drop $\frac{4}{3} \pi R^3=N \times \frac{4}{3} \pi r^3$ or $R^3=N r^3$
or $N=\frac{R^3}{r^3}$,
$\Delta E=\mathrm{T}$ [Initial area of big drop-final area of $N$ drops of radius $(R)$ ]
Now, $\triangle A$ (change in surface area)
$$
=4 \pi R^2-N 4 \pi r^2
$$
$\Delta E($ Energy released $)=\mathrm{T} \times \Delta A$
$\Delta A=4 \pi\left(R^2-N r^2\right)$
$=\mathrm{T} \times 4 \pi\left(R^2-N r^2\right)$
Due to releasing of this energy, the temperature is lowered. If $\rho$ is the density and $s$ is specific heat of liquid and its temperature is lowered by $\Delta t$, then energy released $=m s \Delta t$ [ $s=$ specific heat $\Delta t=$ change in temperature]
So, $\Delta E=m s \Delta t$
Equating both equation,
$$
\begin{aligned}
&\mathrm{T} \times 4 \pi\left(R^2-N r^2\right)=\left(\frac{4}{3} \pi \times R^3 \times \rho\right) s \Delta t \\
&{\left[\because m=V \rho=\frac{4}{3} \pi R^3 \rho\right]} \\
&\Rightarrow \Delta t=\frac{T \times 4 \pi\left(R^2-N r^2\right)}{\frac{4}{3} \pi R^3 \rho \times s}=\frac{3 T}{\rho s}\left[\frac{R^2}{R^3}-\frac{N r^2}{R^3}\right] \\
&=\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{\left(R^3 / r^3\right) \times r^2}{R^3}\right] \quad=\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{1}{r}\right] \\
&
\end{aligned}
$$