B

Step 1: When the disc slides down the inclined plane, it only experiences translational motion, and the acceleration can be calculated using gravitational force components.
Step 2: For the rolling motion, both translational and rotational motions are considered, leading to a different effective acceleration.
Step 3: Using the relation \( t = \sqrt{\frac{2h}{g \sin(\theta)}} \) for sliding and a modified version for rolling, we can find the times \( t_1 \) and \( t_2 \).
Step 4: The ratio is given by \( \frac{t_1}{t_2} = \sqrt{\frac{1}{1 + k^2}} \), where \( k \) is the radius of gyration. For a solid disc, \( k^2 = \frac{1}{2} \), giving us the final ratio.
Therefore, the correct answer is B.