A

Given:
- Radius of the disk, R = 0.2 m
- Force, F = 20 N
- Mass of the disk, M = 20 kg
- Final angular speed, \omega = 50 \text{ rad/s}

Step 1: Calculate the moment of inertia (I) of the disk.
Moment of Inertia of a solid disk about its center:
\[ I = \frac{1}{2} M R^2 = \frac{1}{2} (20)(0.2^2) = 0.4 \text{ kg m}^2 \]

Step 2: Calculate the torque (\tau) applied by the force.
Torque is given by: \[ \tau = F \times R = 20 \times 0.2 = 4 \text{ N m} \]

Step 3: Relate torque to angular acceleration (\alpha).
Using \[ \tau = I \alpha \Rightarrow \alpha = \frac{\tau}{I} = \frac{4}{0.4} = 10 \text{ rad/s}^2 \]

Step 4: Calculate the time (t) taken to reach the final angular speed (\omega) using \[ \omega = \alpha t \Rightarrow t = \frac{\omega}{\alpha} = \frac{50}{10} = 5 \text{ s} \]

Step 5: Calculate the total angle (\theta) covered in radians during this time period.
Using the equation of motion for angular displacement: \[ \theta = \frac{1}{2} \alpha t^2 = \frac{1}{2} (10) (5^2) = 125 ext{ rad} \]

Step 6: Find the number of revolutions (n) the disk completes:
Number of revolutions is given by: \[ n = \frac{\theta}{2\pi} = \frac{125}{2\pi} \approx 19.89 \]
Therefore, to the nearest integer, the number of revolutions n is approximately 20.

Therefore, the answer is approximately: A.