The MI of the thin ring about its diameter,
\(Ir = \frac{1}{2}MR^2 = 1 kg\ m^2\)
Since the ring is melted and recast into a thin disc of same radius R, the mass of the disc equals the mass of the ring = M. The MI of the thin disc about its own axis (i.e. transverse symmetry axis) is
\(I_d= \frac{1}{2}MR^2 = I_r\)
\(∴ I_d = 1 kg\ m^2\)