The correct order of bond orders of C _ 2 ^ 2 - , N _ 2 ^ 2 - and O...

Bond order $= \dfrac{{N}_{b} - {N}_{a}}{2}$

${N}_{B} =$ Number of electrons in bonding molecular orbitals.

${N}_{A \cdot B} =$ Number of electrons in anti bonding molecular orbitals.

${C}_{2}^{2 -}$ has $14$ electrons, it is isoelectronic with ${N}_{2}$ where Bond order $= 3$,

${N}_{2}^{2 -}$ has $16$ electrons is isoelectronic with ${O}_{2}$ whose bond order $= 2$

${O}_{2}^{2 -}$ has $18$ electrons and is isoelectronic with ${F}_{2}$ whose bond order $= 1$

So, the correct order of Bond order is

${C}_{2}^{2 -} > {N}_{2}^{2 -} > {O}_{2}^{2 -}$.