If I₁ is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its center of mass, and I₂ is the moment of inertia (about central axis) of the ring formed by bending the rod, then : 

\(I_1 : I_2 = \pi^2 : 3\)

\(I_1 = \frac{ml^2}{12}\)

\(I_2 = mR^2= m(\frac{l}{2 \pi})^2 \text{Since } l = 2\pi R\) 

  = \(\frac{ml^2}{4 \pi^2}\)

Required ratio : \(\frac{I_1}{I_2} = \frac{ml^2}{12}.\frac{4 \pi^2}{ml^2}\)

  = \(\frac{\pi^2}{3}\)