Consider the pulley shown in the figure, which has a radius of 10...

Let $a$ be the acceleration of the blocks, $T_1$ be the tension in the string attached to the $2.0 \text{ kg}$ block, and $T_2$ be the tension in the string attached to the $4.0 \text{ kg}$ block.

The equations of motion for the blocks along the incline are:
For the $4.0 \text{ kg}$ block (moving downwards):
$m_2 g \sin 45^\circ - T_2 = m_2 a$
$4g \sin 45^\circ - T_2 = 4a$

For the $2.0 \text{ kg}$ block (moving upwards):
$T_1 - m_1 g \sin 45^\circ = m_1 a$
$T_1 - 2g \sin 45^\circ = 2a$

For the rotation of the pulley:
$(T_2 - T_1)R = I\alpha$
Since the string does not slip, $\alpha = \dfrac{a}{R}$.
$\Rightarrow T_2 - T_1 = \dfrac{I}{R^2}a$

Given $I = 0.5 \text{ kg-m}^2$ and $R = 10 \text{ cm} = 0.1 \text{ m}$:
$\dfrac{I}{R^2} = \dfrac{0.5}{(0.1)^2} = \dfrac{0.5}{0.01} = 50 \text{ kg}$

Adding the three equations:
$(4g \sin 45^\circ - T_2) + (T_2 - T_1) + (T_1 - 2g \sin 45^\circ) = 4a + \dfrac{I}{R^2}a + 2a$
$4g \sin 45^\circ - 2g \sin 45^\circ = \left(6 + 50\right)a$
$2g \sin 45^\circ = 56a$
$2g \left(\dfrac{1}{\sqrt{2}}\right) = 56a$
$\sqrt{2}g = 56a$

Taking $g = 10 \text{ m/s}^2$ and $\sqrt{2} \approx 1.414$:
$a = \dfrac{1.414 \times 10}{56} = \dfrac{14.14}{56} \approx 0.25 \text{ m/s}^2$

The acceleration of the $4.0 \text{ kg}$ block is approximately $0.25 \text{ m/s}^2$.

Answer: $0.25 \text{ m/s}^2$