Answer
$I = 4.65~kg~m^2$
Work Step by Step
We can find the block's speed after it drops 12.0 meters.
$\frac{v+v_0}{2} = \frac{12.0~m}{4.00~s}$
$v = 6.00~m/s$
We can find the block's kinetic energy.
$KE = \frac{1}{2}mv^2$
$KE = \frac{1}{2}(8.20~kg)(6.00~m/s)^2$
$KE = 147.6~J$
We can find the angular speed of the wheel.
$\omega = \frac{v}{R} = \frac{6.00~m/s}{0.320~m}$
$\omega = 18.75~rad/s$
We can use conservation of energy to find the moment of inertia of the wheel.
$KE_{block}+KE_{wheel} = mgh$
$\frac{1}{2}I\omega^2 = mgh -KE_{block}$
$I = \frac{2mgh -2KE_{block}}{\omega^2}$
$I = \frac{(2)(8.20~kg)(9.80~m/s^2)(12.0~m) -(2)(147.6~J)}{(18.75~rad/s)^2}$
$I = 4.65~kg~m^2$
