1. For rolling without slipping on an incline

   The linear acceleration of a rigid body rolling down an inclined plane is $a = \frac{g\sin\theta}{1 + \dfrac{I_{cm}}{mR^2}}$

2. Moment of inertia of a solid disc

   For a solid disc about its center, $I_{cm} = \frac{1}{2}mR^2$

   Here, the radius is given as $a$, so $R=a$. Thus, $\frac{I_{cm}}{mR^2} = \frac{\frac12 ma^2}{ma^2} = \frac12$

3. Substitute into the acceleration formula

   $a = \frac{g\sin\theta}{1 + \frac12} = \frac{g\sin\theta}{\frac32} = \frac{2}{3}g\sin\theta$

4. Compare with the given form

   The question says acceleration is $\frac{2}{b}g\sin\theta$

   Comparing, $\frac{2}{b} = \frac{2}{3}$

   Therefore, $b=3$

5. Nearest integer

   $b=3$

Comparison with stored answer:

Stored correct answer = $3$

Derived answer = $3$

They match.