Here, $m = 10\ \text{kg}$; $R = 15\ \text{cm}$ ; $T = 1.5\ \text{s}$ Moment of inertia of the disc about wire, $I = \frac{1}{2} M R^{2}$ $= \frac{1}{2} \times 10 \times (0.15)^{2}$ $=0.1125\ \mathrm{kg}\ \mathrm{m}^{2}$ When the wire is twisted by rotating the disc and then released, restoring torque will be set up. If $\alpha$ is angular acceleration produced, then restoring torque$\tau = I \alpha$...(i)The torsional spring constant k of the wire is defined by the relation$\tau = -k \alpha$...(ii)where θ is the angle of twistFrom the equations (i) and (ii) we have$I \alpha = -k \theta$ or $\alpha = -\frac{k}{I} \theta$Since $\frac{k}{I}$ is constant, it follows that angular acceleration is directly proportional to the angle of twist (angular displacement). Hence, the motion executed by the disc is simple harmonic in nature and the period of torsional oscillation is given by $T = 2\pi \sqrt{\frac{\theta}{\alpha}}$ $=2\pi \sqrt{\frac{\theta}{\frac{k\theta}{I}}} = 2\pi \sqrt{\frac{I}{k}}$or $k = \frac{4\pi^{2} I}{T^{2}}$ $= \frac{4\pi^{2} \times 0.1125}{1.5^{2}} = 1.97\ \mathrm{N}\ \mathrm{m}\ \mathrm{rad}^{-1}$