1. Count total electrons in $O_2^{2-}$

Each oxygen atom has $8$ electrons, so neutral $O_2$ has: $8+8=16 ext{ electrons}$ The charge $2-$ means $2$ extra electrons are added: $16+2=18 ext{ electrons}$

2. Write the molecular orbital filling for $O_2^{2-}$

For $O_2$, $F_2$, and their ions, the MO energy order is: $\sigma(1s)<\sigma^*(1s)<\sigma(2s)<\sigma^*(2s)<\sigma(2p_z)<\pi(2p_x)=\pi(2p_y)<\pi^*(2p_x)=\pi^*(2p_y)<\sigma^*(2p_z)$

Now place $18$ electrons in these orbitals: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,[\pi(2p_x)]^2\,[\pi(2p_y)]^2\,[\pi^*(2p_x)]^2\,[\pi^*(2p_y)]^2$

3. Check for unpaired electrons

All occupied molecular orbitals contain paired electrons. Hence, the number of unpaired electrons is: $0$

4. Final answer

The number of unpaired electron(s) in $O_2^{2-}$ is: $\boxed{0}$

5. Comparison with stored correct answer

Stored correct answer = $0$

My derived answer also = $0$, so they agree.