Answer
$I = 0.976~kg~m^2$
Work Step by Step
We can find the angular acceleration $\alpha$
$\theta = \frac{1}{2}\alpha ~t^2$
$\alpha = \frac{2\theta}{t^2}$
$\alpha = \frac{(2)(8.20~rev)(2\pi~rad/rev)}{(12.0~s)^2}$
$\alpha = 0.7156~rad/s^2$
We can find the angular speed $\omega$ at $t = 12.0~s$:
$\omega = \alpha ~t = (0.7156~rad/s^2)(12.0~s)$
$\omega = 8.587~rad/s$
We can find the moment of inertia $I$.
$K = \frac{1}{2}I\omega^2 = 36.0~J$
$I = \frac{72.0~J}{\omega^2}$
$I = \frac{72.0~J}{(8.587~rad/s)^2}$
$I = 0.976~kg~m^2$
