Answer
$$V_{a}= V_{b } \gt V_c\gt V_{e }= V_{f }$$
Work Step by Step
We know that the electric potential at a point that is between two parallel plates of a capacitor is given by
$$V=Es$$
where $s$ is measured from the negative plate.
Hence, the rank of the electric potentials of the given points are:
$$V_{a}= V_{b } $$
$$V_c$$
$$V_{e }= V_{f } $$
Therefore,
$$\boxed{V_{a}= V_{b } \gt V_c\gt V_{e }= V_{f } }$$