A solid conducting sphere of radius R=20 cm carries a charge Q=+20 nC. It is surrounded by a thin concentric conducting spherical shell of radius 3R…

Heat equals the loss in electrostatic energy. Initially, with inner charge $Q$ (radius $R$) and shell charge $q$ (radius $3R$): $U_i=\dfrac{kQ^2}{2R}+\dfrac{kQq}{3R}+\dfrac{kq^2}{6R}$, where the middle term is the interaction energy of the inner sphere with the shell potential. After joining, all charge $Q+q$ sits on the outer shell (inner sphere is at the same potential and the cavity field vanishes), so $U_f=\dfrac{k(Q+q)^2}{6R}$. Subtracting, $U_i-U_f=\dfrac{kQ^2}{2R}-\dfrac{kQ^2}{6R}=\dfrac{kQ^2}{3R}$, independent of $q$. Numerically, $\dfrac{9\times10^{9}\times(20\times10^{-9})^2}{3\times0.2}=\dfrac{3.6\times10^{-6}}{0.6}=6\times10^{-6}\ \mathrm{J}=6\ \mu\mathrm{J}$. Note why the result is independent of the shell charge: the shell's own self-energy changes only through the total charge, and the cross term $kQq/3R$ is the same before and after, because after joining the whole charge $Q+q$ has the shell's radius. The energy lost appears as Joule heat in the connecting wire (and radiation, neglected here). Using $q=-10\ \mathrm{nC}$ explicitly: $U_i=\dfrac{9\times10^{9}\times4\times10^{-16}}{0.4}-\dfrac{9\times10^{9}\times2\times10^{-16}}{0.6}+\dfrac{9\times10^{9}\times10^{-16}}{1.2}=9-3+0.75=6.75\ \mu\mathrm J$ and $U_f=\dfrac{9\times10^{9}\times10^{-16}}{1.2}=0.75\ \mu\mathrm J$, a loss of $6\ \mu\mathrm J$.