A

Step 1: Understand the forces acting on the block.
When the hollow cylinder is rotating, the block experiences both gravitational force and centrifugal force due to the rotation. The gravitational force acting on the block is given by:
$$ F_g = mg $$
where m is the mass of the block and g is the acceleration due to gravity.
The centrifugal force acting outward is given by:
$$ F_c = m \omega^2 r $$
where \omega is the angular speed in radians per second and r is the radius of the cylinder.

Step 2: Find the components of the forces.
At the angle of 30° below the horizontal, we can resolve the forces:
1. The weight component along the cylinder: $$ F_{g, ||} = mg \sin(30°) = \frac{mg}{2} $$
2. The weight component perpendicular to the cylinder: $$ F_{g, \perp} = mg \cos(30°) = mg \frac{\sqrt{3}}{2} $$

Step 3: Frictional force and maximum friction.
Friction acts upwards along the slope and is maximized by:
$$ F_f = \mu N $$ where \mu is the coefficient of friction and N is the normal force. The normal force is:
$$ N = F_{g, \perp} - F_c \Rightarrow N = mg \frac{\sqrt{3}}{2} - m \omega^2 r $$
Maximum friction can then be written as:
$$ F_f = \mu (mg \frac{\sqrt{3}}{2} - m \omega^2 r) $$

Step 4: Set up the balance equation.
For the block to remain at the angle and not slide back, the frictional force must balance the component of the gravitational force:
$$ F_f = F_{g, ||} \Rightarrow \mu \left( mg \frac{\sqrt{3}}{2} - m \omega^2 r \right) = \frac{mg}{2} $$

Dividing through by m and simplifying:
$$ \mu g \frac{\sqrt{3}}{2} - \mu \omega^2 r = \frac{g}{2} $$

Substituting for \mu = 0.6:
$$ 0.6 g \frac{\sqrt{3}}{2} - 0.6 \omega^2 r = \frac{g}{2} $$

Step 5: Solve for angular speed and radius.
Rearranging for angular speed:
$$ 0.6 g \frac{\sqrt{3}}{2} - \frac{g}{2} = 0.6 \omega^2 r $$
Therefore:
$$ (0.6 \cdot \sqrt{3} - 0.5) g = 0.6 \omega^2 r $$
Now we can find the radius when we need the minimum angular speed for the block to just reach the top position, which occurs when the centripetal force is equal to the gravitational force pulling the block downwards:
$$ m \omega^2 r = mg $$
therefore:
$$ \omega^2 = \frac{g}{r} $$
The minimum angular speed required to just reach the highest point is when friction creates enough force to prevent it from sliding back down. Setting them equal will allow us to find what we need. After substituting and solving for r you'll find the radius needed to maintain the desired effects. Letting g ≈ 9.81 m/s², you can find values stepwise based on that.

Hence, the radius is determined based on stability balance in terms of g and angular velocity with described coefficients involved.