Statement (A): Oxygen molecule has 10 bonding electrons. 

Statement (B): Bond order of ${\mathrm{He}}_2$ molecule is zero.

Statement (C): ${\mathrm{C}}_2$ is diamagnetic in nature.

The molecular orbital electronic configuration of oxygen $\left({\mathrm{O}}_2\right)$molecule is

Where $\mathrm{\sigma },\mathrm{\pi }$, represents bonding molecular orbitals and

${\mathrm{\sigma }}^{*},{\mathrm{\pi }}^{*}$ represents anti bonding molecular orbitals.

So, the total number of electrons present in bonding molecular orbitals of oxygen molecule = 10.

Electronic configuration of ${\mathrm{He}}_2 \mathrm{molecule} \mathrm{is} \mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2$

Bond order$=\frac{1}{2}({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}})=\frac{2-2}{2}=0$

Electronic configuration of ${\mathrm{C}}_2$molecule is 

Bond order$=\frac{1}{2}({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}})=\frac{8-4}{2}=2$

(C = C)

Here all the molecular orbitals of ${\mathrm{C}}_2$ molecule are doubly occupied, so  ${\mathrm{C}}_2$ is diamagnetic in nature.