The bond order can be calculated

as

$\mathrm{B}.\mathrm{O}=\frac{1}{2}\left({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}\right)$

Where, $N_b=$ Electrons in bonding orbitals $N_a=$ Electrons in anti bonding orbitals.

(a) $CO$ and $O_2^{2-}$

The electronic configuration of $CO$ (14) is

$\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\mathrm{\pi }2{\mathrm{p}}^2\mathrm{y}$

$\therefore B.O=\frac{1}{2}(10-4)=\frac{6}{2}=3$

(b) $O_2^{-}$ and $CO$

The electronic configuration of $0_2^{-}\left(17\right)$ is

$\sigma 1s^2{\sigma }^{*}1s^2\sigma 2s^2{\sigma }^{*}2s^2\sigma 2p_z^2\pi 2p_x^2$

$\pi 2p_y^2{\pi }^{*}2p_x^2{\pi }^{*}2p_y^1$

$\mathrm{B}.\mathrm{O}.$ of $\mathrm{CO}$ is $3$

[as calculated in option (a)]

(c) $\mathrm{B}.\mathrm{O}.$. of ${\mathrm{O}}_2^{2-}$ is 1

[as calculated in option (a)]

The electronic configuration of $B_2\left(10\right)$ is

$\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1$

(d) $B .O$. of $C O$ is $3$

[as calculated in option (a)]

Electronic configuration of $N_2^{+}$ ($13$) is

$\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^1$

Thus, option (c) is correct.