The number of moles present in $1{\mathrm{m}}^3$ of any gas at NTP is

22.4 L of any gas contain 1 mole

$\begin{array}{l}\mathrm{or}\frac{22.4}{{10}^3}{\mathrm{m}}^3\mathrm{of}\mathrm{any}\mathrm{gas}\mathrm{contain}1\mathrm{mole}\\ \therefore 1{\mathrm{m}}^3\mathrm{of}\mathrm{any}\mathrm{gas}\mathrm{contain}\frac{{10}^3}{22.4}\mathrm{moles}=44.6\end{array}$