Answer
Please see the work below.
Work Step by Step
For left block
$T=m_1gsin(60)$...........eq(1)
For right block
$T=m_2gsin(20)$...........eq(2)
As the two blocks are in equilibrium, so we can equate the above two equations
$m_1gsin(60)=m_2gsin(20)$
This can be rearranged as:
$\frac{m_1}{m_2}=\frac{gsin(20)}{gsin(60)}$
$\implies \frac{m_1}{m_2}=0.394$
$m_1=0.394m_2$
Thus, the mass of left hand block is 0.394 times the right side block.
