Physics Question?

…
The two masses are equal
M_Al = M_Fe
and M = D V
giving
2.70E3 (2 pi r^3) = 7.89E3 (2 pi 4.80^3)
any conversion factors would just cancel
simplify
2.70 r^3 = 7.86 (4.80^3) = 869.25
r^3 = 321.95
r = 6.85 cm

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Let r cm be the radius of the solid aluminum sphere.

1.00 m³ = (1.00 m)³ = (100 cm)³ = 1.00 × 10⁶ cm³

Density of aluminum = (2.70 × 10³ × 1000 g) / (1.00 × 10⁶ cm³) = 2.70 g/cm³
Density of iron = (7.86 × 10³ × 1000 g) / (1.00 × 10⁶ cm³) = 7.86 g/cm³

Volume of the iron sphere = (4/3) × π × 4.8³ cm³
Mass of the iron sphere = (4/3) × π × 4.8³ × 7.86 g

Volume of the aluminum sphere = (4/3) × π × r³ cm³
Mass of the aluminum sphere = (4/3) × π × r³ × 2.70 g

The aluminum sphere balances the iron sphere on an equal-arm balance. Hence,
Mass of the aluminum sphere = Mass of the iron sphere
(4/3) × π × r³ × 2.70 = (4/3) × π × 4.8³ × 7.86
r = ∛(4.8³ × 7.86 / 2.70)
r = 6.85

Radius of the aluminum sphere = 6.85 cm

Volume V ≡ r^3
r' = r*³√7.86/2.7 = 4.80*1.428 = 6.854 cm

Volume ratio = (7.86/2.7) = 2.9111:1.
Cube root (2.9111 x 4.8^3) = 6.85cm.