A thin ring has a mass of 1.5 kg and a radius of 0.40 m.

Step 1. (a) $I = MR^2 = 0.24\ \text{kg m}^2$.

Step 2. (b) Two diameters at right angles add to (a): $I = \tfrac12 MR^2 = 0.12\ \text{kg m}^2$.

Step 3. (c) Parallel to a diameter, $R$ away: $0.12\ \text{kg m}^2 + 0.24\ \text{kg m}^2 = 0.36\ \text{kg m}^2$.

Step 4. (d) Parallel to the central axis, $R$ away: $0.24\ \text{kg m}^2 + 0.24\ \text{kg m}^2 = 0.48\ \text{kg m}^2$.