Option D is correct. As the number of lone pair of electrons increases, bond angle decreases.
$\mathrm{NO}_{2}^{+}$ ion is isoelectronic with $\mathrm{CO}_{2}$, molecule. It is a linear ion and its central atom $\left(\mathrm{N}}\right)$ undergoes sp-hybridisation. Hence, its bond angle is 180 In $\mathrm{NO}_{2}^{-}$ ion, $\mathrm{N}$ -atom undergoes $\mathrm{sp}^{2}$ hybridisation The angle between hybrid orbital should be 120 but one lone pair of electrons is lying on N-atom hence bond angle decreases to $115^{\circ}$ In $\mathrm{NO}_{2}$ molecule, N-atom has one unpared electron in $s p^{2}$ -hybrid orbital. The bond angle should be $120^{\circ}$ but actually, it is $132^{\circ} .$ it may be due to one unpaired electron in $s p^{2}$ -hybrid orbtal
Therefore, the increasing order of bond angle is
$$
\begin{array}{rrr}
\mathrm{NO}_{2}^{-} & < & \mathrm{NO}_{2} & < & \mathrm{NO}_{2}^{+} \\
115^{\circ} & & 132^{\circ} & & 180^{\circ}
\end{array}
$$