Step-by-Step Solution:

1. Analyze the motion of a general body rolling down an inclined plane. Let the angle of inclination be $\theta$. For a body rolling without slipping, the acceleration of its center of mass is given by the formula: $a = \frac{g \sin\theta}{1 + \frac{I}{MR^2}}$ where $I$ is the moment of inertia, $M$ is the mass, and $R$ is the radius of the body.

2. Calculate the acceleration for the ring. For a ring, the moment of inertia is $I_{ring} = MR^2$. Substituting this into the acceleration formula: $a_{ring} = \frac{g \sin\theta}{1 + \frac{MR^2}{MR^2}} = \frac{g \sin\theta}{1 + 1} = \frac{g \sin\theta}{2}$

3. Calculate the acceleration for the disc. For a disc, the moment of inertia is $I_{disc} = \frac{1}{2}MR^2$. Substituting this into the acceleration formula: $a_{disc} = \frac{g \sin\theta}{1 + \frac{\frac{1}{2}MR^2}{MR^2}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2g \sin\theta}{3}$

4. Relate travel time to acceleration and height. Let $h$ be the height of the inclined plane. The length of the path along the incline is $s = \frac{h}{\sin\theta}$. The objects start from rest, so the initial velocity $u=0$. Using the kinematic equation $s = ut + \frac{1}{2}at^2$, we get: $s = \frac{1}{2}at^2 \implies t = \sqrt{\frac{2s}{a}}$ Substituting $s = \frac{h}{\sin\theta}$: $t = \sqrt{\frac{2h}{a \sin\theta}}$

5. Calculate the time taken for the ring and the disc. For the ring: $t_{ring} = \sqrt{\frac{2h}{a_{ring} \sin\theta}} = \sqrt{\frac{2h}{\left(\frac{g \sin\theta}{2}\right) \sin\theta}} = \sqrt{\frac{4h}{g \sin^2\theta}} = \frac{2}{\sin\theta}\sqrt{\frac{h}{g}}$ For the disc: $t_{disc} = \sqrt{\frac{2h}{a_{disc} \sin\theta}} = \sqrt{\frac{2h}{\left(\frac{2g \sin\theta}{3}\right) \sin\theta}} = \sqrt{\frac{3h}{g \sin^2\theta}} = \frac{\sqrt{3}}{\sin\theta}\sqrt{\frac{h}{g}}$

6. Set up and solve the equation for the time difference. Since $a_{ring} < a_{disc}$, the ring takes more time to reach the bottom, so $t_{ring} > t_{disc}$. The time difference is: $\Delta t = t_{ring} - t_{disc} = \frac{2}{\sin\theta}\sqrt{\frac{h}{g}} - \frac{\sqrt{3}}{\sin\theta}\sqrt{\frac{h}{g}} = \frac{2 - \sqrt{3}}{\sin\theta}\sqrt{\frac{h}{g}}$ We are given $\theta = 60^\circ$, $g = 10 \, m/s^2$, and $\Delta t = \frac{2 - \sqrt{3}}{\sqrt{10}} \, s$. We also know $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Substituting these values into the equation: $\frac{2 - \sqrt{3}}{\sqrt{10}} = \frac{2 - \sqrt{3}}{\frac{\sqrt{3}}{2}} \sqrt{\frac{h}{10}}$ The term $(2 - \sqrt{3})$ cancels out from both sides: $\frac{1}{\sqrt{10}} = \frac{2}{\sqrt{3}} \sqrt{\frac{h}{10}}$ The term $\sqrt{10}$ also cancels out: $1 = \frac{2}{\sqrt{3}} \sqrt{h}$ Solving for $\sqrt{h}$: $\sqrt{h} = \frac{\sqrt{3}}{2}$ Squaring both sides to find $h$: $h = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} = 0.75 \, m$

Thus, the height of the top of the inclined plane is 0.75 metres.