Mass of man m = 70 kg
Radius of the cylindrical drum r = 3 m
Frequency of the vertical axis f = 200 rev/min
Coefficient of the friction \mu = 0.15

Step 1: Note down the forces acting on the man
Frictional force acting upward is balanced by the weight of the man acting down ward
f_r = mg .....................(1)
The normal force acting towards the central axis gives centripetal force.
N = F_c
N = \frac{mv^2}{r} = mr \omega^2 ........(2)
Step 2: Set up an equation for the minimum angular velocity
Frictional force F_r = \mu * N
f_r = \mu * (mr \omega^2) \because \text{ equation (2) }
mg = \mu * (mr \omega^2) \because \text{ equation (1) }
g = \mu * (r \omega^2)
\omega = \sqrt{\frac{g}{\mu * r}}
Step 3: Find the minimum angular speed at which the drum rotates
\omega = \sqrt{\frac{10}{0.15 * 3}}
\omega = 4.7 rad/sec
Hence, minimum angular speed at which the drum rotates \omega = 4.7 rad/sec