Answer
$I = 0.079~kg \cdot m^2$
Work Step by Step
We can find the period:
$T = \frac{50~s}{20~cycles} = 2.5~s$
We can find the rotational inertia:
$T = 2\pi \sqrt{\frac{I}{\kappa}}$
$\frac{T}{2\pi} = \sqrt{\frac{I}{\kappa}}$
$\frac{T^2}{4\pi^2} = \frac{I}{\kappa}$
$I = \frac{T^2\kappa}{4\pi^2}$
$I = \frac{(2.5~s)^2 (0.50 ~N\cdot m)}{4\pi^2}$
$I = 0.079~kg \cdot m^2$
