Answer
The percent of the iceberg that is above the water surface is 10.5%
Work Step by Step
Let $V_i$ be the total volume of the iceberg. Let $V_w$ be the volume of the iceberg that is submerged in the water. Note that $V_w$ is the volume of water that is displaced. The buoyant force is equal to the weight of the iceberg. Therefore;
$F_B = M_i~g$
$\rho_w~V_w~g = \rho_i~V_i~g$
$\frac{V_w}{V_i} = \frac{\rho_i}{\rho_w}$
$\frac{V_w}{V_i} = \frac{0.917 \times 10^3~kg/m^3}{1.025\times 10^3~kg/m^3}$
$\frac{V_w}{V_i} = 0.895 = 89.5\%$
The percent of the iceberg that is submerged is 89.5%. Therefore, the percent of the iceberg that is above the water surface is 10.5%.