Answer
a) $2$
b) $\dfrac{1}{4}$
c) $1$
Work Step by Step
We know that the electric field inside an ideal capacitor is a uniform electric field and is given by
$$E=\dfrac{\eta}{\epsilon_0}=\dfrac{Q}{A\epsilon_0}$$
$$\color{blue}{\bf [a]}$$
when $Q_f=2Q$, while $A$ is constant,
$$\dfrac{E_f}{E}=\dfrac{\dfrac{Q_f}{A\epsilon_0}}{\dfrac{Q}{A\epsilon_0}}=\dfrac{Q_f}{Q}=\dfrac{2Q}{Q}$$
Thus,
$$\dfrac{E_f}{E}=\color{red}{\bf 2}$$
$$\color{blue}{\bf [b]}$$
when $L_f=2L$, where $A=L^2$ and hence, $A_f= (2L)^2=4L^2$. and $Q$ is constant,
$$\dfrac{E_f}{E}=\dfrac{\dfrac{Q}{A_f\epsilon_0}}{\dfrac{Q}{A\epsilon_0}}=\dfrac{A}{A_f}=\dfrac{L^2}{4L^2}$$
Thus,
$$\dfrac{E_f}{E}=\color{red}{\bf \dfrac{1}{4}}$$
$$\color{blue}{\bf [c]}$$
When $d_f=2d$, nothing happens to the final electric field since the electric field between the plates is independent of $d$. So that
$$\dfrac{E_f}{E}=\color{red}{\bf1}$$
