= Solution
The closed surface integral sums the outward electric <flux>: $\mathbf E$ is the <electric field>, $d\mathbf S$ its outward vector <area>, $Q$ the total enclosed charge and $\epsilon_0$ the <vacuum> permittivity. <Gauss's law> relates that <flux> to enclosed charge for any closed surface. High <symmetry> can make the field's magnitude constant on a suitable <Gaussian surface> and fix its direction; without <symmetry>, <flux> alone does not determine the local field.
For an isolated conducting <sphere>, charges reside at its surface and the exterior field is radial. Thus
$$
\boxed{\mathbf E(r)=\frac{Q}{4\pi\epsilon_0r^2}\widehat{\mathbf r}\quad(r>a),\qquad
V(a)=\int_a^\infty E(r)\,dr=\frac{Q}{4\pi\epsilon_0a},\qquad C=4\pi\epsilon_0a.}
$$
The conductor has zero interior field. Integrating exterior field <energy> gives
$$
U=\int_a^\infty\frac12\epsilon_0\left(\frac{Q}{4\pi\epsilon_0r^2}\right)^2 4\pi r^2dr
=\boxed{\frac{Q^2}{8\pi\epsilon_0a}}.
$$
Let the initial charge per drop be $q=4\pi\epsilon_0aV_1$. After disconnecting the charging supply, charge is conserved. <Volume> conservation gives final <radius> $A=2^{1/3}a$, total charge $2q$ and
$$
\boxed{V_f=\frac{2q}{4\pi\epsilon_0A}=2^{2/3}V_1.}
$$
Initial mutual <energy> is negligible at the stipulated large separation. Hence
$$
U_i=\frac{q^2}{4\pi\epsilon_0a},\qquad
U_f=\frac{4q^2}{8\pi\epsilon_0\,2^{1/3}a}=2^{2/3}U_i,\qquad
\boxed{U_i/U_f=2^{-2/3}.}
$$
\b[The <electrostatic energy increase on equal-drop coalescence> is positive.] It can be supplied by the decrease in surface <energy> during coalescence and/or by external mechanical work bringing like charges together. Remaining <energy> can become motion and heat. If the charge is too large, surface-energy release alone need not suffice to make merging spontaneous; conservation does not assert that it will. Keeping the drops connected to a <voltage> supply would instead be a different charge-balance problem.
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