Answer
(a) $V_1 = $ 4240 V
(b)$V_2 = $ 8480 V
(c) $W = 8.3 \times 10^{-3} \,\text{J}$
Work Step by Step
(a) The potential between the two plates of the capacitor is related to the charges on the plates and the capacitance in the form
$$V_1 = \dfrac{Q}{C} = \dfrac{3.9 \times 10^{-6} \,\text{C}}{920 \times 10^{-12} \,\text{F}} = \boxed{4240 \,\text{V} }$$
(b) The voltage is directly proportional to the separated distance, therefore, as the distance is doubled, the potential will be doubled and becomes
$$V_{2} = 2 \times 4240 \,\text{V} = \boxed{8480 \,\text{V}}$$
(c) The work done to double the separation equals the difference between the energies between the two instants and is given by
\begin{gather*}
W = \dfrac{1}{2} QV_2 - \dfrac{1}{2} QV_1\\
W = \dfrac{1}{2} Q( V_2 - V_1)\\
W = \dfrac{1}{2} (3.9 \times 10^{-6} \,\text{C}) ( 8480 \,\text{V} - 4240\,\text{V}) \\
\boxed{W = 8.3 \times 10^{-3} \,\text{J}}
\end{gather*}
