Given the area is:
A=4.0 cmx4.0 cm
The seperation distance is d=5.0 mm
The capacitance is :
[tex]\begin{gathered} C=\frac{\epsilon_oA}{d} \\ \Rightarrow C=8.85\times10^{-12}\frac{4.0\times10^{-2}\times4.0\times10^{-2}}{5\times10^{-3}} \\ \Rightarrow C=2.83\times10^{-12}F \end{gathered}[/tex]Thus the answer is
[tex]2.8\times10^{-12}F[/tex]