Number of atoms $=\frac{\text { mass }}{\text { molar mass }} \times N_{A}$
$\times$ Number of atoms in $1$ mole
$\therefore $ Number of atoms in $4.25 g NH _{3}$
$=\frac{4.25}{17} \times N_{A} \times 4 $
$=N_{A}$
Number of atoms in $8 \,g$
$O _{2}=\frac{8}{32} \times N_{A} \times 2=\frac{N_{A}}{2}$
Number of atoms in $2\, g$
$H _{2}=\frac{2}{2} \times N_{A} \times 2=2 N_{A}$
Number of atoms in $4\, g$
$He =\frac{4}{4} \times N_{A} \times 1=N_{A}$
Thus, $2 \,g\, H _{2}$ contains the maximum number of atoms among the given.