Answer: $ \large\frac{m_2g}{m_1+m_2}$

We can draw the free body diagrams for our problem as follows:

When the system is released, let the acceleration of the blocks be aa.

From the Free Body Diagrams, for the bigger block $m_1 \rightarrow T = m_1a$ and for the lighter block $m_2 \rightarrow m_2g - T = m_2a$

Solving for $a$, we get $a = \large\frac{m_2g}{m_1+m_2}$