Why not use permittivity of free space with insulator between capacitor plates

[bonbon22]

Homework Statement

A capacitor of capacitance 7.2pF consists of two parallel metal plates separated by an
insulator of thickness 1.2mm. The area of overlap between the plates is 4.0 × 10−4m2.
Calculate the permittivity of the insulator between the capacitor plates.

Relevant Equations

capacitance = Eo Er A /d

the answer is 2.16*10^-11
what i don' t understand about the question is why did they not use the constant Eo 8.85418782 × 10-12
in this case as there is a insulator occupying the area inbetween does that mean then their is no free space ??
Also in this equation the variable A is the area of each capacitor plate or both together ?
They mentioned "area of overall" but I am not sure what they mean by this.

thank you if you reply.

 

[Charles Link]

$\epsilon_r$ is the relative permittivity, but they are looking for the absolute permittivity which is the product $\epsilon_r \epsilon_o$. $\\$ For the area, it is simply the area of one face plate. $\\$ And the entire region between the plates is filled with a dielectric material=(like a plastic, etc). If you compute $\epsilon_r$, I think you get something like 2.4, which is reasonable.