A string is wound around a wheel having a moment of inertia of 0.20...

Let $a$ be the downward acceleration of the block and $T$ be the tension in the string.

For the block of mass $m$, the equation of motion is:
$mg - T = ma$

For the wheel, the tension $T$ provides the torque for its rotation. The equation of rotational motion is:
$\tau = I\alpha$
$TR = I\alpha$

Since the string does not slip, the linear acceleration $a$ is related to the angular acceleration $\alpha$ by:
$a = \alpha R \Rightarrow \alpha = \dfrac{a}{R}$

Substituting $\alpha$ in the torque equation:
$TR = I\left(\dfrac{a}{R}\right)$
$T = \dfrac{Ia}{R^2}$

Substituting the expression for $T$ into the block's equation of motion:
$mg - \dfrac{Ia}{R^2} = ma$
$mg = a\left(m + \dfrac{I}{R^2}\right)$
$a = \dfrac{mg}{m + \dfrac{I}{R^2}}$

Given values are:
$m = 2.0 \text{ kg}$
$g = 9.8 \text{ m/s}^2$
$I = 0.20 \text{ kg-m}^2$
$R = 10 \text{ cm} = 0.1 \text{ m}$

Calculating the equivalent mass of the wheel:
$\dfrac{I}{R^2} = \dfrac{0.20}{(0.1)^2} = \dfrac{0.20}{0.01} = 20 \text{ kg}$

Now, substituting the values to find $a$:
$a = \dfrac{2.0 \times 9.8}{2.0 + 20}$
$a = \dfrac{19.6}{22}$
$a \approx 0.89 \text{ m/s}^2$