The bond dissociation energy (E) and bond length $\left(\mathrm{R}\right)$ of ${\mathrm{O}}_2,{ \mathrm{N}}_2,{ \mathrm{F}}_2$ follows the order as

${\mathrm{N}}_2$ has triple bond between $\mathrm{N}$ atoms.
${\mathrm{O}}_2$ has double bond between O atoms.
${\mathrm{F}}_2$ has single bond between the atoms.
The bond energy will be higher, if the bonds between the atoms is high.
In ${\mathrm{F}}_2$, the atoms are closer and repulsions are more. So, bond energy of ${\mathrm{F}}_2$ is less.
$\therefore$ Order of bond energies $={\mathrm{N}}_2>{\mathrm{O}}_2>{\mathrm{F}}_2$

Bond length is ${\mathrm{N}}_2<{\mathrm{O}}_2<{\mathrm{F}}_2$