1. Write the moments of inertia about a diameter

For equal mass $M$ and respective radii:

• Solid disc about a diameter: $I_d = \frac{1}{4}MR_d^2$

• Ring about a diameter: $I_r = \frac{1}{2}MR_r^2$

• Solid sphere about a diameter: $I_s = \frac{2}{5}MR_s^2$

Since the sphere has the same radius as the disc, $R_s = R_d$

2. Use the given condition for disc and ring

The disc's moment of inertia is $2.5$ times that of the ring: $I_d = 2.5\, I_r$

Substitute the formulas: $\frac{1}{4}MR_d^2 = 2.5\left(\frac{1}{2}MR_r^2\right)$

Cancel $M$: $\frac{1}{4}R_d^2 = \frac{5}{4}R_r^2$

Multiply by $4$: $R_d^2 = 5R_r^2$

So, $\frac{R_d^2}{R_r^2} = 5$

3. Find the sphere's moment of inertia relative to the ring

We need $n$ such that $I_s = n I_r$

Now, $n = \frac{I_s}{I_r} = \frac{\frac{2}{5}MR_s^2}{\frac{1}{2}MR_r^2}$

Since $R_s = R_d$, $n = \frac{\frac{2}{5}MR_d^2}{\frac{1}{2}MR_r^2} = \frac{2}{5}\cdot 2 \cdot \frac{R_d^2}{R_r^2}$

$n = \frac{4}{5}\cdot 5 = 4$

4. Final answer

$\boxed{n=4}$

The derived answer matches the stored correct answer.