Assuming conservation of
volume, we find the radius of the combined spheres, then use $$C =
4\pi\varepsilon_0R$$ to find the capacitance. When the drops combine, the
volume is doubled. It is then $$V = 2(4\pi/3)R^3$$. The new radius $$R'$$ is
given by :
$$\dfrac{4\pi}{3}(R')^3=2\dfrac{4\pi}{3}R^3\Rightarrow
R'=2^{1/3}R$$
The new capacitance is :
$$C'=4\pi\varepsilon_0R'=4\pi\varepsilon_02^{1/3}R=5.04\pi\varepsilon_0R$$
With $$ R=2.00mm,$$ we obtain $$
C=5.04\pi(8.85\times10^{-12}F/m)(2.00\times10^{-3}m)=2.80\times10^{-13}F.$$