The acceleration of a body rolling down an inclined plane without slipping is given by the formula $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$, where $k$ is the radius of gyration and $R$ is the radius of the body. For a solid sphere, the moment of inertia is $I = \frac{2}{5}MR^2$, so $\frac{k^2}{R^2} = \frac{2}{5} = 0.4$. For a uniform disc, the moment of inertia is $I = \frac{1}{2}MR^2$, so $\frac{k^2}{R^2} = \frac{1}{2} = 0.5$. Since $\frac{k^2}{R^2}$ is smaller for the solid sphere, its acceleration $a$ will be greater than that of the disc ($a_{\text{sphere}} > a_{\text{disc}}$). The object with greater acceleration will take less time to cover the same distance and reach the bottom first. Also, the acceleration is independent of the mass of the rolling body. Therefore, the solid sphere will get to the bottom first.