The total number of atoms in $4.25g$ of $N{H}_3$ is approximately:

The correct option is C $6\times {10}^{23}$
Moles of $N{H}_3=\frac{4.25}{17}=0.25$

Number of atoms in 0.25 moles of $N{H}_3$ is $0.25\times 6.023\times {10}^{23}\times 4=6.023\times {10}^{23}$