When the iceberg sits in the water it displaces a volume of water equal to its own weight. Imagine a perfect cube of solid ice. It floats in the water with its top surface a distance x above the water line. The weight of the ice is volume times density, so since the volume is 1 unit, the weight is 0.92 units. The volume of water displaced is 1-x units because the distance from the base of the iceberg to the surface of the water is 1-x. Therefore the weight of the water displaced is 1.03(1-x). This has to balance the weight of the iceberg=0.92, so 1.03(1-x)=0.92, and 1.03-1.03x=0.92. 1.03x=1.03-0.92=0.11, so x=0.11/1.03=0.1068 approximately, or 10.7%. As a formula we can put w as the density of water and i as the density of ice. So w(1-x)=i, and w-wx=i, so x=(w-i)/w. Since we had a unit cube, x is actually the fraction of the volume of the iceberg above water. This of course does not take into account the shape of the iceberg.