The capacitance of the capacitor is 
Explanation:
The capacitance of a parallel-plate capacitor is given by the equation

where
k is the dielectric constant of the medium
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
For the capacitor in this problem, we have:
k = 2.1 is the dielectric constant
is the separation between the plates
(I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)
Therefore, the capacitance of the capacitor is

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