Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
$\textbf{When the Mylar sheet is inside the capacitor:}$
We know that the capacitance of a parallel plate capacitor is given by
$$C=\kappa C_0=\dfrac{\kappa \epsilon_0 A}{d} $$
where $\kappa$ is the dielectric constant.
Plug the known;
$$C= \dfrac{\kappa_{\rm Mylar} \epsilon_0 A}{d} =\dfrac{(3.1)(8.85\times 10^{-12})(5\times 5\times 10^{-6})}{(0.10\times 10^{-3})}$$
$$C= {\bf 6.86\times 10^{-12}}\;\rm F$$
Since the battery is connected for enough time to the capacitor, the capacitor must have the same potential difference as the battery.
$$\Delta V_C=V_{B}=\color{red}{\bf 9}\;\rm V$$
Recalling that $\Delta V_C=Q/C$. So the charge inside the capacitor is given by
$$Q=\Delta V_C C=(9)(6.86\times 10^{-12})$$
$$Q=\color{red}{\bf 6.17\times 10^{-11} }\;\rm C$$
We know that the electric field inside the capacitor is given by
$$E=\dfrac{\Delta V_C}{\kappa d}=\dfrac{(9)}{(3.1)(0.1\times 10^{-3})}$$
$$E=\color{red}{\bf 2.9\times 10^4}\;\rm V/m$$
$$\color{blue}{\bf [b]}$$
$\textbf{When the Mylar sheet is withdrawn from the capacitor:}$
$$C=\kappa C_0=\dfrac{\kappa \epsilon_0 A}{d} $$
where $\kappa$ is the dielectric constant.
Plug the known;
$$C= \dfrac{\kappa_{\rm vacuum} \epsilon_0 A}{d} =\dfrac{(1)(8.85\times 10^{-12})(5\times 5\times 10^{-6})}{(0.10\times 10^{-3})}$$
$$C= {\bf 2.21\times 10^{-12}}\;\rm F$$
Since the battery is still connected to the capacitor, the capacitor must have the same potential difference as the battery.
$$\Delta V_C=V_{B}=\color{red}{\bf 9}\;\rm V$$
Recalling that $\Delta V_C=Q/C$. So the charge inside the capacitor is given by
$$Q=\Delta V_C C=(9)(2.21\times 10^{-12})$$
$$Q=\color{red}{\bf 1.99\times 10^{-11} }\;\rm C$$
We know that the electric field inside the capacitor is given by
$$E=\dfrac{\Delta V_C}{\kappa_{\rm vacuum} d}=\dfrac{(9)}{(1)(0.1\times 10^{-3})}$$
$$E=\color{red}{\bf 9.0\times 10^4}\;\rm V/m$$
