Answer
The magnitude of the electron's acceleration is the same in all three situations.
Work Step by Step
The electric field due to the positively charged sheet is directed down and the electric field due to the negatively charged sheet is directed down. Then the net electric field is the sum of the magnitudes of the electric fields due to the two sheets.
The magnitude of the force on the electron is $F = \vert q \vert E$, and the magnitude of the acceleration is $a = \frac{F}{m}$
Thus, the magnitude of the acceleration is proportional to the magnitude of the net electric field.
We can find the net electric field in each situation.
Situation 1:
$E_{net} = \frac{4\sigma}{2~\epsilon_0}+ \frac{4\sigma}{2~\epsilon_0} = \frac{8\sigma}{2~\epsilon_0}$
Situation 2:
$E_{net} = \frac{7\sigma}{2~\epsilon_0}+ \frac{\sigma}{2~\epsilon_0} = \frac{8\sigma}{2~\epsilon_0}$
Situation 3:
$E_{net} = \frac{3\sigma}{2~\epsilon_0}+ \frac{5\sigma}{2~\epsilon_0} = \frac{8\sigma}{2~\epsilon_0}$
Since the net electric field is the same in all three situations, the magnitude of the electron's acceleration is the same in all three situations.
