College Physics (7th Edition)

$2.16\times10^{-9}\,F$
Given/known: $\kappa=1$ (as air is in between the plates), $A=0.525\,m^{2},$ $d=2.15\,mm=2.15\times10^{-3}\,m$ and $\varepsilon_{0}=8.854\times10^{-12}\,F/m$ Recall: $C=\frac{\kappa\varepsilon_{0}A}{d}$ Result: $C=\frac{1\times8.854\times10^{-12}\,F/m\times0.525\,m^{2}}{2.15\times10^{-3}\,m}=2.16\times10^{-9}\,F$