In 7 g nitrogen, number of molecules $ = \frac {7.0} {28} mol$
$ = 0.25 \times N_A$ molecules
where, $N_A = \, Avogadro \, number \, = 6.023 \times 10^{23}$
In 2 g of $H_2 = \frac {2.0}{2} mol \, = 1 \times N_A$ molecules
In 16 g of \, $NO_2 = \frac {16.0} {46} mol$
$ \, \, \, \, \, \, \, \, = 0.348 \times N_A$ molecules
In $16\, g$ of $O_2 = \frac {16}{32} \, mol \, = 0.5 \times N_A $ molecules
Hence, maximum number of molecules are present in $2\, g$ of $H_2$.