Answer
a) $10$
b) $1$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The initial charge density of the shape is $\eta_i=Q/A_i$ and let's assume that $\eta_f=Q/A_f$ is the final charge density of the shape when it is compressed in the $x-y$ plane.
This means that the area of the shape changes from $A_i$ to $A_f$ where
$$A_f=\dfrac{A_i}{3.163^2}\tag 1$$
Hence,
$$\dfrac{\eta_f}{\eta_i}=\dfrac{\dfrac{ \color{red}{\bf\not} Q}{A_f}}{\dfrac{ \color{red}{\bf\not} Q}{A_i}} =\dfrac{A_i}{A_f} $$
Plugging from (1);
$$\dfrac{\eta_f}{\eta_i}= \dfrac{ \color{red}{\bf\not} A_i}{\dfrac{ \color{red}{\bf\not} A_i}{3.163^2}}= {{3.163^2}}$$
$$\dfrac{\eta_f}{\eta_i}=\color{red}{\bf 10}$$
$$\color{blue}{\bf [b]}$$
Since the electron is far away, it perceives the object as a point charge with a net charge of $Q$ in both cases: whether the shape is reduced or remains constant.
Thus,
$$\dfrac{E_f}{E_i}=\color{red}{\bf 1}$$
