The acceleration of a rigid body rolling down an inclined plane without slipping is given by $a = \frac{g \sin\theta}{1 + \frac{K^2}{R^2}}$, where $K$ is the radius of gyration. The ratio $\frac{K^2}{R^2}$ for the given bodies are:

1. Solid sphere: $\frac{K^2}{R^2} = \frac{2}{5} = 0.4$
2. Disc: $\frac{K^2}{R^2} = \frac{1}{2} = 0.5$
3. Solid cylinder: $\frac{K^2}{R^2} = \frac{1}{2} = 0.5$ Since the solid sphere has the smallest value of $\frac{K^2}{R^2}$, it will have the maximum acceleration ($a$) among the three. Therefore, it takes the minimum time and reaches the bottom first.