The oxidation number of ‘$N$’ in $N{H}_3$ is
 

It is given to find out the oxidation state of $N$in $N{H}_3$.

Oxidation state of $H$ is $+1$

Let oxidation state of $N$ be$x$

Overall charge on $N{H}_3$ is $0$

Then,

$$\begin{gathered} x + 3 (+1) = 0 \\ x + 3 = 0 \\ x = -3 \end{gathered}$$