#### Answer

The length of steel that is above the surface is 8.4 cm

#### Work Step by Step

Let $V_c$ be the volume of the cylinder. We can find the mass of the cylinder as:
$M = V_c~\rho_{steel}$
$M = (\pi)(0.05~m)^2(0.20~m)(7900~kg/m^3)$
$M = 12.4~kg$
The buoyant force on the cylinder is equal to the cylinder's weight $Mg$. The buoyant force is equal to the weight of the mercury that is displaced. Let $\rho_m$ be the density of mercury. We can find the volume $V_m$ of mercury that is displaced. Therefore;
$F_B = Mg$
$\rho_m~V_m~g = Mg$
$V_m = \frac{M}{\rho_m}$
$V_m = \frac{12.4~kg}{13.6\times 10^3~kg/m^3}$
$V_m = 9.12\times 10^{-4}~m^3$
We can find the height $h$ such that the bottom part of the cylinder which is submerged has the volume $V_m$ as:
$\pi~r^2~h = V_m$
$h = \frac{V_m}{\pi~r^2}$
$h = \frac{9.12\times 10^{-4}~m^3}{(\pi)~(0.05~m)^2}$
$h = 0.116~m = 11.6~cm$
The length of steel that is below the surface is 11.6 cm. Therefore, the length of steel that is above the surface is 20 cm - 11.6 cm which is 8.4 cm.