Answer
$8.106\times10^{-5}\;kg.m^2$
Work Step by Step
For the torsion pendulum
$T=2\pi\sqrt {\frac{I}{\kappa}}$
or, $I=\frac{\kappa T^2}{4\pi^2}$
From the given graph, $T = 0.40\;s$ and $\kappa=\frac{4\times10^{-3}}{0.2}\;N.m/rad=0.02\;N.m/rad$.
Therefore, the rotational inertia is
$I=\frac{0.02\times 0.40^2}{4\times\pi^2}\;kg.m^2$
or, $\boxed{I=8.106\times10^{-5}\;kg.m^2}$
