The correct order of $\mathrm{N}-\mathrm{O}$bond length in $\mathrm{NO},{\mathrm{NO}}_2^{-}$ and ${\mathrm{NO}}_3^{-}$ will be

Nitrogen atom has 2+5=7 electrons
Oxygen atom has 2+6=8 electrons
Now NO has 6 electrons in the bonding and 1 in the antibonding.
${\mathrm{NO}}_2^{+}$has 2 electrons lesser than NO; So 6 electrons in the bonding and no electrons in the antibonding.

$$\begin{gathered} \text{ Thus, }{\mathrm{NO}}_2^{+}\text{, Bond Order }=\frac{6-0}{2}=3 \\ \text{ For }{\mathrm{NO}}_2^{-},\text{Band order }=\frac{6-3}{2}=1.5 \\ \text{ For }{\mathrm{NO}}_3^{-}\text{, Band order }=\frac{6-4}{2}=1 \\ \text{ The order of Band order is : } \\ {\mathrm{NO}}_3^{-}>{\mathrm{NO}}_2^{-}>{\mathrm{NO}}_2^{+} \end{gathered}$$