Answer
$I = 0.0138~kg~m^2$
Work Step by Step
In part (b), we found that $~~T_2 = 4.87~N$
In part (c), we found that $~~T_1 = 4.54~N$
In part (d), we found that $~~\alpha = 1.20~rad/s^2$
We can find the rotational inertia of the pulley:
$\tau = I \alpha$
$(T_2-T_1)~r= I \alpha$
$I = \frac{(T_2-T_1)~r}{\alpha}$
$I = \frac{(4.87~N-4.54~N)~(0.0500~m)}{1.20~rad/s^2}$
$I = 0.0138~kg~m^2$
