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.
$$\color{blue}{\bf [b]}$$
The battery is still connected 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 unchanged, but the stored charge on each plate will do.
$$\Delta V_{f}=V_B=\color{red}{\bf 9}\;\rm V$$
Hence,
$$Q_f=C_f\Delta V_f=C_fV_B$$
where $C_f=\dfrac{\epsilon_0 A}{d_f}$
Hence,
$$Q_f =\dfrac{\epsilon_0 AV_B}{d_f} $$
Plug the known;
$$Q_f =\dfrac{(8.85\times 10^{-12})(0.02^2)(9)}{ (2\times 10^{-3})}$$
$$Q_f=\pm\color{red}{\bf 1.593\times 10^{-11}}\;\rm C$$
