The rod is pivoted at a distance 3L from one end. The centre of mass is at a distance 6L from the pivot.
Using energy conservation:
mg6ℓ=21Iω2
Moment of inertia about the pivot (using parallel axis theorem):
I=12mℓ2+m(6ℓ)2=12mℓ2+36mℓ2=9mℓ2
mg6ℓ=21⋅9mℓ2⋅ω2
ω2=6⋅2ℓ9g=ℓ3g
ω=ℓ3g