Answer
See the detailed answer below.
Work Step by Step
The given figure represents two capacitors, one on the left and one on the right.
We can assume that the electric field between the plates [inside the capacitors] is uniform and constant.
We know that the electric field inside a capacitor is given by
$$E=\dfrac{\eta}{\epsilon_0}=\dfrac{Q}{A\epsilon_0}\tag 1$$
and we know that the electric potential is given by
$$V=V_i-\int Edx$$
and since $E$ is uniform inside the capacitor,
$$V=V_i- E \Delta x\tag 2$$
$$\color{blue}{\bf [a]}$$
At $x=0.5$ cm,
From (1)
$$E= \dfrac{50\times 10^{-9}}{(0.02)^2(8.85\times 10^{-12})} $$
Thus,
$$\vec E=-(\color{red}{\bf 1.41\times 10^7}\;{\rm V/m})\;\hat i$$
The negative sign is due to the direction of the electric field, as you see in the figure below.
From (2),
$$V=0- (-1 .41\times 10^7)(0.5-0)\times 10^{-2}$$
$$V=\color{red}{\bf 70.5}\;\rm kV$$
$$\color{blue}{\bf [b]}$$
At $x=1.5$ cm,
We know that the electric field inside a conductor is zero.
$$\vec E=\color{red}{\bf }0\;{\rm V/m} $$
However, the electric potential will be equal to the electric potential on the surface of this conductor (plate), which is at $x=1$ cm.
From (2),
$$V=0- (-1 .41\times 10^7)(1-0)\times 10^{-2}$$
$$V=\color{red}{\bf 141}\;\rm kV$$
$$\color{blue}{\bf [c]}$$
At $x=2.5$ cm,
From (1)
$$E= \dfrac{50\times 10^{-9}}{(0.02)^2(8.85\times 10^{-12})} $$
Thus,
$$\vec E= (\color{red}{\bf 1.41\times 10^7}\;{\rm V/m})\;\hat i$$
The direction of the electric field is to the right.
From (2),
$$V=0- ( 1 .41\times 10^7)(-0.5-0)\times 10^{-2}$$
We assumed that the last right plate [the negative plate] is the zero potential here. So $x=-0.5$ cm here.
Also by symmetry, the electric field at this point is equal in magnitude to the first point but opposite in direction. So the electric potential must be the same.
$$V=\color{red}{\bf 70.5}\;\rm kV$$
