Paramagnetic molecules are those molecules in which unpaired electrons are present, while diamagnetic molecules are those in which unpaired electrons are absent.

The diatomic species with an odd number of electrons and those diatomic species that contain 10 or 16 electrons are paramagnetic. Among the given species ${\mathrm{O}}_2^{+}\left(15\right)$ and $B_2\left(10\right)$ are paramgnetic species.
The molecular orbital electronic configuration of ${\mathrm{B}}_2$ is.

${\mathrm{B}}_2:\mathrm{\sigma }1{\mathrm{s}}^2, {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1$

Bond order$=\frac{\mathrm{Bonding} \mathrm{electrons}-\mathrm{antibonding} \mathrm{electrons}}{2}$

$=\frac{6-4}{2}=\frac{2}{2}=1$

Hence. the answer is $1$