With no friction, block 1 would be pulled to the edge of the table as block 2 falls. By Newton's second law, the net horizontal force on block 1 is
∑ F[1, h] = T = (2 kg) a
where T = magnitude of tension and a = acceleration of the blocks, while the net force on block 2 is
∑ F[2] = (3 kg) g - T = (3 kg) a
where (3 kg) g = weight of block 2 and g = 9.8 m/s². Notice that we take the downward fall of block 2 to be the positive direction.
Eliminate T and solve for a :
T + ((3 kg) g - T) = (2 kg) a + (3 kg) a
(3 kg) g = (5 kg) a
a = 3g/5
Solve for T :
T = (2 kg) (3g/5) = 6g/5 kg = 11.76 N
Now, taking friction into account, block 1 would have net force
∑ F[1, h] = T - sf = 0
so that the maximum magnitude of static friction is sf = 11.76 N. The net vertical force on block 1 is
∑ F[1, v] = n - (2 kg) g = 0
where n = magnitude of the normal force. Let µ be the coefficient of static friction. It follows that
n = (2 kg) g = 19.6 N
sf = µ n ⇒ µ = (11.76 N) / (19.6 N) = 0.6