Buoyant force and finding the weight of water required to float object 
Find the weight of water required to float the pontoon in the casting dock below when the gap all around is $0.1 \text{ m}$. The total weight the pontoon is $500 \text{ t}$ ($1 \text{ t} = 8896 \text{ N}$).

Is this consistent with Archimedes principle, i.e. does upthrust equal weight of the displaced fluid?

The pressure on the base to support the weight is
$$F_b = \rho \cdot g \cdot z$$
Using vertical equilibrium, $F_b = 4.448 \text{ MN}$, therefore $z = 2.83 \text{ m}$. My confusion lies in the next part. See below a solution from my lecturer. I am really confused at what I am looking at, especially finding the volume of water. Is the volume of water we are after the volume of water before the block has been placed?

 A: The first Volumen he calculates is the actual volume of water that you can see in your little figure (or atleast a cross section).
The second volume is the amount of water that you would need to replace the part of the pontoon that is under water. Meaning, if you took out the pontoon and then filled the tank up to height h = 2.934m.
This amount of water that is replaced by the floating object $O$ is of interest because from the equation of equal forces you get:
$m_{O}g = \rho_{H_2O}gz_rA = \rho_{H_2O}gV_r $
$\rho_OV_O = \rho_{H_2O}V_r$
$\frac{\rho_O}{\rho_{H_2O}} = \frac{V_r}{V_O}$
Here the index $r$ is for the water that is replaced by floating object.
In your case you an cancel the surface $A$ that makes up the Volume $V_r=z_rA$ and you get an equation for the height instead of volumes.
Actually I think there is a little error in your eqaution (or at least notation) since $F$ is usually a force, but $\rho g h $ is a pressure (?).
Edit: As R.W Bird notices: seems that your Prof forgets that there should also be 0.1m of water gap at the top and bottom of the pontoon. I will calculate it as four sides and the bottom:
$V = (2\cdot2.834\cdot0.1\cdot(40.2+4)+4.2\cdot0.1\cdot40.2) m^3 = 41.94 m^3$
A: Assuming the weight of the pontoon = the weight of water displaced, then (500 tons)(8896 Nt/ton) = (1000 kg/$m^3$)((4)(40)(z)$m^3$)(9.8 Nt/kg), and z = 2.8367 m, in good agreement with your value.  However, the volume of water in the doc must be divided into five segments: the bottom and each of four sides. I got 72.978 $m^3$. This has no relation to the required displacement. The pontoon is supported by the fluid pressure on its bottom, which depends only on the submersion depth. Consider that the pontoon was initially floating outside of the dock (and displacing water).  It was then floated into the dock,  and the  gate was closed, excluding some of the displaced water.
