Volume of a mixture of $6.02\times {10}^{23}$ oxygen atoms and $3.01\times {10}^{23}$ hydrogen molecules at $NTP$ is:

The correct option is D $22.4litre$
$6.02\times {10}^{23}$ oxygen atoms = 1 mole of Oxygen atom
$n_1=0.5$ mole of $O_2$

$3.01\times {10}^{23}$ hydrogen molecules, $n_2=0.5$ mole of $H_2$
$n_1+n_2=0.5+0.5=1$

Total volume of 1 mole of gases $=1\times 22.4=22.4L$
Option D is correct.