Answer
(a) $d$ = 1.1 mm
(b) $V_{ab}$ = 84 V
Work Step by Step
(a) The capacitance between the two plates depends on the separated distance between the two plates and we could use equation 24.2 in the textbook to determine the distance $d$ in the form
\begin{equation}
d = \dfrac{\epsilon_o A V_{ab}}{Q} \tag{1}
\end{equation}
Let us substitute the values of $\epsilon_o, A, V_{ab}$ and $Q$ into equation (1) to get the value of $d$
\begin{align}
d = \dfrac{\epsilon_o A V_{ab}}{Q} = \dfrac{(8.85 \times 10^{-12} \,\text{F/m}) (6.8 \times 10^{-4} \mathrm{~m^2}) (42 \,\text{V})}{240\times 10^{-12} \,\text{C}} = 1.1 \,\text{mm}
\end{align}
(b) The voltage is directly proportional to the separated distance, therefore, as the distance is doubled, the potential will be doubled to be
$$V_{ab} = 2 \times 42 \,\text{V} = 84 \,\text{V}$$
