Answer
$\alpha = 4.6~rad/s^2$
Work Step by Step
Let $T$ be the tension in the cord.
We can consider the forces on the block to write an expression for the tension $T$:
$P - T = m~a$
$T = P - m~a$
We can consider the torque on the wheel to find the angular acceleration of the wheel:
$\tau = I~\alpha$
$T~R = I~\alpha$
$(P - m~a)~R = I~\alpha$
$(P - m~\alpha~R)~R = I~\alpha$
$P~R - m~\alpha~R^2 = I~\alpha$
$P~R = I~\alpha+m~\alpha~R^2$
$\alpha = \frac{P~R}{I+m~R^2}$
$\alpha = \frac{(3.0~N)(0.20~m)}{(0.050~kg~m^2)+(2.0~kg)~(0.20~m)^2}$
$\alpha = 4.6~rad/s^2$
