Answer
$I = 1.6~kg~m^2$
Work Step by Step
We can find the angular acceleration of the device:
$\alpha = \frac{a}{r}$
$\alpha = \frac{0.80~m/s^2}{0.20~m}$
$\alpha = 4.0~rad/s^2$
We can find the tension in the rope pulling up on the box:
$T-mg = ma$
$T = mg+ma$
$T = m~(g+a)$
$T = (30~kg)~(9.8~m/s^2+0.80~m/s^2)$
$T = 318~N$
We can find the rotational inertia of the device:
$\sum~\tau = I~\alpha$
$R~F_{app} - r~T = I~\alpha$
$I = \frac{R~F_{app} - r~T}{\alpha}$
$I = \frac{(0.50~m)(140~N) -(0.20~m)(318~N)}{4.0~rad/s^2}$
$I = 1.6~kg~m^2$
