Answer:
at the beginning: 
when the plates are pulled apart: 
Explanation:
The capacitance of a parallel-plate capacitor is given by

where
k is the relative permittivity of the medium (for air, k=1, so we can omit it)
is the permittivity of free space
A is the area of the plates of the capacitor
d is the separation between the plates
In this problem, we have:
is the area of the plates
is the separation between the plates at the beginning
Substituting into the formula, we find

Later, the plates are pulled apart to
, so the capacitance becomes
