Hi there!
Assuming the pulley is frictionless and massless, we can do a summation of forces for each block.
Block 1 has the tension force (in direction of acceleration) and force of static friction acting on it. For the block to be stationary, the net force must be 0, so:
[tex]\Sigma F = T - F_S[/tex]
[tex]0 = T - F_S\\\\T = F_S[/tex]
Now, we can do a summation for Block 2.
[tex]\Sigma F = W_2 - T\\\\0 = W_2 - T\\\\T = W_2[/tex]
We can substitute this tension in the above equation to solve for the force of static friction and the coefficient of static friction.
[tex]F_S = W_2\\\\\mu m_1g = m_2g\\\\\mu = \frac{m_2}{m_1} = \boxed{\mu = .2}[/tex]
