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, right after the battery is disconnected, the potential difference of the capacitor is still 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, the charge of the plates remained constant and so did the electric field inside the capacitor.
Hence,
$$E_2=\color{red}{\bf 3000}\;\rm V/m$$
$$Q_2=\color{red}{\bf2.1\times 10^{-10}}\;\rm C$$
But the potential difference changes,
$$(\Delta V_C)_2=Ed_2=3000(1\times 10^{-2})$$
$$(\Delta V_C)_2=\color{red}{\bf 30}\;\rm V$$
$$\color{blue}{\bf [c]}$$
Now the charge is still unchanged but the area did change so the electric field will be changed as a result.
$$Q_3=\color{red}{\bf2.1\times 10^{-10}}\;\rm C$$
$$E_3= \dfrac{Q_3}{A_3\epsilon_0}$$
where $A_3=\pi r_3^2$ and $r_3=2r_1$, $\Rightarrow$ $A_3=4\pi r_1^2=4A_1$
$$E_3= \dfrac{Q_3}{4A_1\epsilon_0}=\dfrac{E_1}{4}=\dfrac{3000}{4}$$
$$E_3=\color{red}{\bf 750}\;\rm V/m$$
Hence the potential difference changes too,
$$(\Delta V_C)_3=E_3d_1=(750)(0.5\times 10^{-2})$$
$$(\Delta V_C)_3=\color{red}{\bf 3.75}\;\rm V$$
