A small metal sphere of radius 1.0 cm is used to charge a large metal sphere of radius 20 cm, which…

Step 1. One charging (a). $q = \dfrac{rV_0}{k} = \dfrac{0.010\ \text{m} \times 3.0 \times 10^{3}\ \text{V}}{9.0 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}} = 3.33\ \text{nC}$.

Step 2. One sharing (b). Joined, the spheres reach one potential, so the charge divides in the ratio $R : r = 20 : 1$. The large sphere keeps $\dfrac{20}{21}$ of it: $Q_1 = 3.17\ \text{nC}$.

Step 3. The rule (c). $Q_n = \dfrac{R}{R + r}\,(Q_{n-1} + q)$. Its steady value is $Q_\infty = \dfrac{qR}{r} = 66.67\ \text{nC}$, and $Q_n = Q_\infty\left[1 - \left(\dfrac{R}{R + r}\right)^n\right]$. After two cycles $Q_2 = 6.198\ \text{nC}$, so $V_2 = \dfrac{kQ_2}{R} = 279\ \text{V}$.

Step 4. The limit (d). $V_\infty = \dfrac{kQ_\infty}{R} = \dfrac{kq}{r} = V_0 = 3.00\ \text{kV}$. This makes sense: once the large sphere is at $V_0$, sharing moves no more charge.

Step 5. Cycles (e). We need $\left(\dfrac{20}{21}\right)^n \le 0.1$, so $n \ge \dfrac{\ln 10}{\ln(21/20)} = 47.19$. The first whole number is $48$.

Step 6. Heat (f). In the end both spheres sit at $V_0$. The supply has sent out all the charge they hold, $\dfrac{(R + r)V_0}{k}$, and done work $\dfrac{(R + r)V_0^2}{k}$. They store only half of that, $\dfrac{(R + r)V_0^2}{2k}$. The other half is heat, made in both steps of every cycle: in the supply's wire while the small sphere is topped up, and in the long wire while the charge is shared. In all it is $\dfrac{0.21\ \text{m} \times (3.0 \times 10^{3}\ \text{V})^2}{2 \times 9.0 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}} = 105\ \mu\text{J}$.