Answer: Option C
:
C
Moment of inertia of rod AB about its centre and perpendicular to the length $=\frac{m{l}^2}{12}=l\therefore m{l}^2=12l$
Now moment of inertia of the rod about the axis which is passing through O and perpendicular to the plane of hexagon $l_{rod}=\frac{m{l}^2}{12}+m{x}^2$ [From the theorem of parallel axes]
$=\frac{m{l}^2}{12}+m{\left(\frac{\sqrt{3}}{2}l\right)}^2=\frac{5m{l}^2}{6}$
Now the moment of inertia of system $l_{system}=6\times I_{rod}=6\times \frac{5m{l}^2}{6}=5m{l}^2$
$l_{system}=5\left(12l\right)=60l$ [As $m{l}^2=12l]$