Let the radius of bigger droplet is $R$, then by volume conservation:

$\left(\frac{4}{3}\pi R^3\right)=1000\left(\frac{4}{3}\pi {\left(1\right)}^3\right)$

$\Rightarrow R=10 \mathrm{mm}$

Final potential energy:

$U_f=1000\left(4\pi R^{2}T\right)=1000\times 4\pi {\left(10\times {10}^{-3}\right)}^{2}T$

Initial potential energy:

$U_i=4\pi \times {\left(r\right)}^{2}T=4\pi \times {\left({10}^{-3}\right)}^{2}T$

Therefore, change in potential energy,

$∆\mathrm{E}=T\times 1000\times 4\mathrm{\pi }{\left({10}^{-3}\right)}^2-\mathrm{T}\times 4\mathrm{\pi }{\left(10\times {10}^{-3}\right)}^2$

$∆\mathrm{E}=4\times \mathrm{\pi }\times 7\times {10}^{-2}\left[1000-100\right]\times {10}^{-6}$

$∆\mathrm{E}=7.92\times {10}^{-4}\mathrm{J}$