Let `rho` and `rho_(1)` be the density of ice and see water respectively.
so, `rho//rho_(1)=0.9`
Let H be the thickness of ice block, of crosso section area A.
The weight of ice block = `A H rho g`
Let x be the thickness of iceblock above the surface of water
`:.` Thickness of ice block inside the water
`=(H-x)`
By the principle of floatation,
`AH rhog=(H-x) A rho_(1) g`
or `x=H(1-rho//rho_(1)) = 5(1-0.9)=0.5 m`.