An ice cube with a small iron ball in its centre is placed in a cup of water. 3.9 x 10-3kg of water in the cup is displaced and the volume of the ice cube is 4.0 x 10-6m3. Ice density: 1000 kg m-3 Iron density: 7800 kg m-3, what is the volume of the iron?

Firstly you need to know that the weight/mass of water displaced by the cube is equal to the weight/mass of the ice cube: 3.9 x 10^-3kg You then need to recognise that the weight of the cube is the weight of the iron ball added to the weight of the ice.  Where the volume of the iron ball is V, the weight of the iron ball is the volume of the iron ball multiplied by the density iron: V x 7800 kg m^-3.
The weight of the ice will be the volume of the ice (the total volume of the ice cube minus the volume of the iron) multiplied by the density of iron: (4 x 10^-6m³– V) x 920 kg m^-3 Hence mass of the ice cube = mass of ice + mass of iron
3.9 x 10^-3= ((4 x 10^-6– V) x 920) + (V x 7800) 3.9 x 10^-3= (3.68 x 10^-3-920V) + 7800V (3.9 x 10^-3) – (3.68 x 10^-3) = 6880V 2.2 x 10^-4= 6880V V = 3.2 x 10^-8 Volume of iron ball = 3.2 x 10^-8m³