Answer
The acceleration of the 3.0-kg block is $\frac{2m_2~g}{m_1+4m_2}$.
Work Step by Step
We can set up a force equation for the 1.0-kg block. Let $m_2$ be the mass of this block. Let $T$ be the tension in the rope. So;
$\sum F = m_2~a_2$
$m_2~g - T = m_2~a_2$
$T = m_2~g - m_2~a_2$
We can use this expression for the tension $T$ in the force equation for 3.0-kg block. Let $m_1$ be the mass of this block. Note that $a_2 = 2a_1$. Then;
$\sum F = m_1~a_1$
$2T = m_1~a_1$
$2m_2~g - 2m_2~a_2 = m_1~a_1$
$2m_2~g - 2m_2~(2~a_1) = m_1~a_1$
$2m_2~g = m_1~a_1+4m_2~a_1$
$a_1 = \frac{2m_2~g}{m_1+4m_2}$
The acceleration of the 3.0-kg block is $\frac{2m_2~g}{m_1+4m_2}$.
