The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to

The expression for capacitance of a parallel plate capacitor is given by $\mathrm{C}=\frac{{\mathrm{ϵ}}_0\mathrm{A}}{\mathrm{d}}$
Where ϵ₀ is the permittivity of free space, A is area of each plate, d is separation between the plates.
So, C$\propto$A
C∝1/d 
Here, it is given that area A of each plate is halved, separation becomes doubled, so capacitance becomes C/4.