Answer
$a = \frac{m_2~g}{m_2 + m_1 + \frac{I}{R^2}}$
Work Step by Step
Let $T_1$ be the tension in the rope attached to block $m_1$. We can find an expression for $T_1$:
$T_1 = m_1~a$
Let $T_2$ be the tension in the rope attached to block $m_2$. We can find an expression for $T_2$:
$m_2~g - T_2 = m_2~a$
$T_2 = m_2~g - m_2~a$
Let $R$ be the radius of the pulley. We can consider the torque from the tension in each rope acting on the pulley:
$\tau_2-\tau_1 = I~\alpha$
$R~T_2 - R~T_1 = I~(\frac{a}{R})$
$(m_2~g-m_2~a) - (m_1~a) = I~(\frac{a}{R^2})$
$m_2~g = m_2~a + m_1~a + \frac{Ia}{R^2}$
$a = \frac{m_2~g}{m_2 + m_1 + \frac{I}{R^2}}$
