Answer
See the detailed answer below.
Work Step by Step
The battery will charge the two plates, one will build up a positive charge and the other will build a negative charge until the potential difference between the two plates equals the potential difference of the battery. See the figure below.
$$\color{blue}{\bf [a]}$$
The battery is connected for enough time until there is no charge going in or out of the two parallel plates,
So the potential difference between the two plates is given by
$$\Delta V_{i}=V_B=\color{red}{\bf 9}\;\rm V$$
The two parallel plates are forming a capacitor where its initial capacitance is given by
$$C_i=\dfrac{\epsilon_0 A}{d_i}$$
the area of the two plates is constant but the distance between the two plates will change in part b.
Plug the known
$$C_i=\dfrac{(8.85\times 10^{-12})(0.02^2)}{(1\times 10^{-3})}=\bf 3.54\times 10^{-12}\;\rm F$$
Recalling that
$$\Delta V_C=\dfrac{Q}{C}$$
Thus,
$$Q_i=C_i\Delta V_i$$
Plug the known;
$$Q_i=(3.54\times 10^{-12})(9)$$
$$Q_i=\pm\color{red}{\bf 3.186 \times 10^{-11}}\;\rm C$$
The plus and minus signs are for the two plates since one of them will be fully charged positively while the other is fully charged negatively.
$$\color{blue}{\bf [a]}$$
The battery now is removed and the two plates are moved away from each other an extra 1 mm. This means that the potential difference between the two plates is changed but not the stored charge on its plate.
The charge is conserved and constant since the plates are still insulated.
Hence,
$$Q_f=Q_i=\pm\color{red}{\bf 3.186 \times 10^{-11}}\;\rm C$$
$$\Delta V_f=\dfrac{Q}{C_f}$$
where $C_f=\dfrac{\epsilon_0 A}{d_f}$
$$\Delta V_f=\dfrac{Qd_f}{\epsilon_0 A}$$
Plug the known;
$$\Delta V_f=\dfrac{(3.186 \times 10^{-11})(2\times 10^{-3})}{(8.85\times 10^{-12})(0.02^2)}$$
$$\Delta V_f=\color{red}{\bf 18}\;\rm V$$
