The pair of species with the same bond order is :

According to molcular orbital theory.
$O_2^{2-}$ (8+8+2=18)
$=\sigma 1s^2,\stackrel{\star }{\sigma }1s^2,\sigma 2s^2,\stackrel{\star }{\sigma }2s^2,\sigma 2s_z^2,\pi 2p_x^2\approx \pi 2p_y^2,\stackrel{\star }{\pi }2p_x^2\approx \stackrel{\star }{\pi }2p_y^2$
Bond order (BC)=$\frac{N_b-N_a}{2}=\frac{10-8}{2}=1$
$B_2$(5+5=10)=$\sigma 1s^2,\stackrel{\star }{\sigma }1s^2,\sigma 2s^2,\stackrel{\star }{\sigma }2s^2,\pi 2p_x^1\approx \pi 2p_y^1$
BO=$\frac{6-4}{2}=1$
Thus, $O_2^{2-}$ and $B_2$ have the same bond order .
NOTE
BO of $O_2^{+}=2.5,N{O}^{+}=3,NO=2.5,CO=3,N_2=3\text{and}{O}_2=2$