Answer
(a) $r_b$ = 17.5 cm
(b) $Q= 25.5 \times 10^{-9} \,\text{C} $
Work Step by Step
(a) The capacitance for a spherical capacitor is calculated by
\begin{gather*}
C = 4 \pi \epsilon_o \dfrac{r_a r_b}{r_b - r_a} \tag{Solve for $r_b$}\\
r_b = \dfrac{r_a C}{C - 4 \pi \epsilon_o r_a}
\end{gather*}
Substitute to get $r_b$
\begin{align*}
r_b &= \dfrac{r_a C}{C - 4 \pi \epsilon_o r_a}\\
&= \dfrac{(0.15 \,\text{m}) (116 \times 10^{-12} \,\text{F})}{116 \times 10^{-12} \,\text{F} - 4 \pi (8.85 \times 10^{-12} \,\text{F/m}) (0.15 \,\text{m})}\\
&= 17.5 \,\text{cm}
\end{align*}
(b) The charges on each sphere are given by
$$Q = CV$$
Let us substitute the values $C$ and $V$ to get the charge on each sphere
\begin{align*}
Q &= CV \\
&= (116 \times 10^{-12} \,\text{F})(220 \,\text{V})\\
&= 25.525 \times 10^{-9} \,\text{C}
\end{align*}
