= Solution
For <charge sharing between distant connected spheres>, use the standard approximation that the spheres are sufficiently far apart for mutual electrostatic effects to be negligible, and that the wire adds negligible <capacitance>. The <capacitance of an isolated conducting sphere> is $C=4\pi\epsilon_0r$. Thus
$$
C_1=5.56\,\mathrm{pF},\qquad C_2=11.1\,\mathrm{pF}=2C_1.
$$
The initial charges are $Q_i=C_iV_i$. Connecting the spheres allows charge to flow until their potentials are equal. <Charge conservation> gives
$$
V_f=\frac{C_1V_1+C_2V_2}{C_1+C_2}
=\frac{(2\,\mathrm{kV})+2(4\,\mathrm{kV})}{3}
=3.333\,\mathrm{kV}.
$$
The initial and final <electrostatic energies> are $U_i=\tfrac12C_1V_1^2+\tfrac12C_2V_2^2$ and $U_f=\tfrac12(C_1+C_2)V_f^2$. Their difference becomes <Joule heating> in the resistive wire:
$$
Q_{\rm heat}=U_i-U_f=\frac12\frac{C_1C_2}{C_1+C_2}(V_1-V_2)^2.
$$
Numerically,
$$
\boxed{Q_{\rm heat}=\frac12\left(3.71\times10^{-12}\,\mathrm F\right)(2000\,\mathrm V)^2=7.42\times10^{-6}\,\mathrm J=7.42\,\mu\mathrm J.}
$$
The resistance determines how fast this happens, not the final <energy> difference. Sufficient damping suppresses sustained charge-current oscillations; the calculation also neglects <energy> radiated away. At finite sphere separation the <capacitance matrix> would be needed, and the radii and initial potentials alone would not specify an exact correction.
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