Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric field inside a capacitor is uniform and is given by
$$E=\dfrac{\Delta V_C}{d}=\dfrac{Q}{A\epsilon_0}\tag1$$
After a long time of charging the capacitor by connecting it to the battery, both, the battery and the capacitor, must have the same potential difference of $V_B=\Delta V_C=15$ V.
Thus, after a long time, the potential difference of the capacitor is 15 V.
$$(\Delta V_C)_1=\color{red}{\bf 15}\;\rm V$$
So that, from (1),
$$E=\dfrac{15}{0.5\times 10^{-2}}$$
$$E_1=\color{red}{\bf 3000}\;\rm V/m$$
And hence, from (1) too,
$$Q=EA\epsilon_0=(3000)\pi (5\times 10^{-2})^2 (8.85\times 10^{-12})$$
$$Q_1=\color{red}{\bf2.1\times 10^{-10}}\;\rm C$$
$$\color{blue}{\bf [b]}$$
If we pulled the electrodes away from each other until they were 1.0 cm apart, by using the insulating handles, while the battery was still connected, then the potential difference remains constant.
$$(\Delta V_C)_2=\color{red}{\bf 15}\;\rm V$$
$$E_2=\dfrac{(\Delta V_C)_2}{d_2}=\dfrac{15}{1\times 10^{-2}}$$
$$E_2=\color{red}{\bf 1500}\;\rm V/m$$
$$Q_2=E_2A\epsilon_0=(1500)\pi (5\times 10^{-2})^2 (8.85\times 10^{-12})$$
$$Q_2=\color{red}{\bf1.04\times 10^{-10}}\;\rm C$$
$$\color{blue}{\bf [c]}$$
Now the potential difference is still unchanged.
$$(\Delta V_C)_3=\color{red}{\bf 15}\;\rm V$$
And hence,
$$E_3=\dfrac{(\Delta V_C)_3}{d_1}=\dfrac{15}{0.5\times 10^{-2}}$$
$$E_3=\color{red}{\bf 3000}\;\rm V/m$$
$$Q_3=E_3A_3\epsilon_0 =(3000)\pi (10\times 10^{-2})^2 (8.85\times 10^{-12})$$
$$Q_3=\color{red}{\bf 8.34\times 10^{-10}}\;\rm C$$
