Answer
$1.1\times10^{27}$ air molecules.
1900 moles.
Work Step by Step
Assume the air is an ideal gas, and that the pressure is 1.0 atm.
$$PV=NkT$$
Find the number of air molecules, N.
$$N=\frac{PV}{kT}=\frac{(1.013\times10^5 Pa)( 6.0\times 3.0 \times 2.5 m^3)}{(1.38\times10^{-23}J/K)(295K) }\approx 1.1\times10^{27}$$
Now find the number of moles.
$$(1.1197\times10^{27})(\frac{1\;mole}{6.02\times10^{23}})=1900\;moles$$
