v = 3.77m/1.6s
mgh = (1/2)mv^2 + (1/2)*I*omega^2
omega = v/R
(20kg)(9.8)(3.77m) = (1/2)(20kg)(2.36 m/s)^2 + (1/2)I((2.36)^2/(0.60)^2)
I = 88.3 ... which is more than I = MR^2 = 25.2 ...

So this is entirely wrong, because I know it should be less, since the mass is more evenly distributed.
Second approach, I believe the drops 3.77 m in 1.6 s is still due to gravity, is not constant velocity. There must also be tension then since it is less than gravitational acceleration.

mg - T = ma
y=(1/2)a*t^2
(3.77m) = (1/2)a(1.6)^2
a= 2.95 m/s^2

(20 kg)(9.8) - T = (20kg)(2.95)
T = 137N

torque = F*R
Only T does something in relation to the rotation of the axis? So:
torque = T*R = I * alpha
alpha = a/R = 2.95m/s^2 / 0.60m = 4.92 rad/s^2

137N * 0.60m = I * (4.92 rad/s^2)
I = 16.6 N s

Well I got the less inertia part now. But was that the right way to do it?

Both ways should give you the same result. You made an error in your first calculation. A hint as to where that error is can be found in your second calculation (a = 2.95m/s²). How fast is the mass moving at the end of 1.6 seconds?