Hollow section replacing solid compression member

Condition: the hollow member must carry the same axial load as the solid one with the same material and the same stress, so the cross-sectional area must be equal. Let the hollow section have outer diameter $D$ and inner diameter $d_i = 0.8D$.

$$A_{solid} = \frac{\pi}{4}(50)^2 = 1963.5\ \text{mm}^2$$ $$A_{hollow} = \frac{\pi}{4}(D^2 - d_i^2) = \frac{\pi}{4}D^2(1-0.8^2) = \frac{\pi}{4}D^2(0.36)$$

Equating areas:

$$0.36D^2 = 50^2 \Rightarrow D = \frac{50}{0.6} = 83.33\ \text{mm}, \qquad d_i = 0.8D = 66.67\ \text{mm}$$

Wall thickness $t = \dfrac{D - d_i}{2} = 8.33$ mm.

Comparison (why the hollow section is better in compression)

For the same weight, the hollow section has a much larger moment of inertia and radius of gyration, so its slenderness ratio is smaller and it resists buckling better.

Answer: outer diameter $\approx 83.3$ mm and inner diameter $\approx 66.7$ mm.