Answer
$I = 0.60~kg~m^2$
Work Step by Step
In part (a), we found that the magnitude of the frictional force is $F_f = 4.0~N$
To find the rotational inertia of the wheel about the rotation axis through its center of mass, we can consider the torque about this axis. Note that only the frictional force produces a torque about this rotation axis.
We can find the rotational inertia of the wheel:
$\tau = I~\alpha$
$R~F_f = I~(\frac{a}{R})$
$I = \frac{R^2~F_f}{a}$
$I = \frac{(0.30~m)^2~(4.0~N)}{0.60~m/s^2}$
$I = 0.60~kg~m^2$
