Answer
$\frac{V_{ice}}{V_{sub}}=\frac{515}{476}$
Work Step by Step
Please see the attached image first.
We can write,
The iceberg's weight is $W_{ice}=m_{ice}g=\rho_{ice}V_{ice}g$
The weight of the displaced water is $W_{water}=m_{water}g=\rho_{water}V_{sub}g$
By Archimedes' principle, $W_{water}$ is equal in magnitude to the buoyancy force, which balances gravity when the iceberg is in equilibrium. So,
$\rho_{water}V_{sub}g=\rho_{ice}V_{ice}g$
$\frac{\rho_{water}}{\rho_{ice}}=\frac{V_{ice}}{V_{sub}}$ ; Let's plug known values into this equation.
$\frac{1030\space kg/m^{3}}{952\space kg/m^{3}}=\frac{V_{ice}}{V_{sub}}$
$\frac{515}{476}=\frac{V_{ice}}{V_{sub}}$
