First let's get the symbols right:
m1 = larger mass, 4 kg
m2 = smaller mass, 2 kg
T1 = rope tension on the m1 side
T2 = rope tension on the m2 side
r = radius of the pulley, 0.085 m
mp = mass of pulley, 2.6 kg
M = moment or torque due to friction at pulley spindle, 0.51 Nm
J = polar moment of inertia of pulley, 0.5 mp r^2 = ... kg-m^2
a = constant acceleration of the system, till m1 hits ground
Now then, let's start;
at the m1 side, gravity is causing tension T1 after catering to acceleration a, so:
T1 = m1 g - m1 a .........
at the pulley, moment of the tension difference at the center caters to both the frictional torque as well as the angular acceleration of the pulley, which is a/r, hence:
(T1 - T2) r = J a/r + M, or
T1 - T2 = J a/r^2 + M /r = 0.5 mp a + M /r ... 2)
at the m2 side, the torque T2 is simply pulling up mass m2 against gravity at acceleration a, hence:
T2 = m2 g + m2 a ......... 3)
Now, for solving, the simplest way is to eliminate the tensions and get a simple equation in a only, as we are not interested in the other unknowns. The best way is to subtract 3) from 1) and equate that to 2) to eliminate T1 and T2
Once you find a, it's a piece of cake to find the time required to go down a height of 1 m, starting from rest with the equation:
h = 0.5 a t^2, or t = sqrt (2 h/a)
1.58 m/s, 1.78 s