Which of the following species exhibits the diamagnetic behaviour?

Atoms or molecules in which all electrons are paired are called diamagnetic; the presence of even a single unpaired electron makes the species paramagnetic. Hence we must count the total electrons in each species and fill them into the molecular-orbital (MO) energy levels to see whether any remain unpaired.

For second-row diatomic molecules the MO sequence after $$1s$$ and $$2s$$ orbitals is
$$\sigma_{2p_z}\;(2e) \lt \pi_{2p_x}=\pi_{2p_y}\;(4e) \lt \pi^{*}_{2p_x}=\pi^{*}_{2p_y}\;(4e) \lt \sigma^{*}_{2p_z}\;(2e).$$
We therefore fill electrons in the order
$$\sigma 1s,\;\sigma^{*}1s,\;\sigma 2s,\;\sigma^{*}2s,\;\sigma 2p_z,\;\pi 2p_x=\pi 2p_y,\;\pi^{*}2p_x=\pi^{*}2p_y,\;\sigma^{*}2p_z.$$ Only the $$2p$$‐based orbitals decide magnetism, because the lower $$1s$$ and $$2s$$ MOs are always filled with paired electrons.

Electrons in 2p MOs = $$16 - 8 = 8$$.
Filling order:
$$\sigma_{2p_z}^2\;\pi_{2p_x}^2\;\pi_{2p_y}^2\;\pi^{*}_{2p_x}^1\;\pi^{*}_{2p_y}^1$$
Two unpaired electrons reside in the degenerate $$\pi^{*}$$ orbitals ⇒ paramagnetic.

Total 2p electrons = $$15 - 8 = 7$$.
Configuration:
$$\sigma_{2p_z}^2\;\pi_{2p_x}^2\;\pi_{2p_y}^2\;\pi^{*}_{2p_x}^1$$
One unpaired electron ⇒ paramagnetic.

Total 2p electrons = $$18 - 8 = 10$$.
Configuration:
$$\sigma_{2p_z}^2\;\pi_{2p_x}^2\;\pi_{2p_y}^2\;\pi^{*}_{2p_x}^2\;\pi^{*}_{2p_y}^2$$
All electrons are paired ⇒ diamagnetic.

The MO ordering for heteronuclear NO is very similar to that of $$O_2^{+}$$. With 15 electrons there is one unpaired electron in a $$\pi^{*}$$ orbital ⇒ paramagnetic.

Thus, among the given species, only $$O_2^{2-}$$ has all its electrons paired and therefore shows diamagnetism.

Option A which is: $$O_2^{2-}$$