Number of atoms in $4.25g$ of ${NH}_3$ is approx:

The correct option is D $6\times {10}^{23}$
Moles of $N{H}_3=\frac{4.25}{17}=0.236$ moles

$\therefore$ Number of molecules $=0.236\times 6.022\times {10}^{23}$ molecules

$1$ molecule of $N{H}_3$ contains $4$ atoms

$\therefore$ Number of atoms=$4\times 0.236\times 6.022\times {10}^{23}$

$\approx 6\times {10}^{23}$ atoms

Hence, option $D$ is correct.