Think of it this way. If the tank was filled to the height of 2.934 m and a drain tube was installed that would empty all water above that level, you would find that the volume of water that passed through the drain tube is equal to the weight of the pontoon or 453.44 m^3.
The volume of water remaining in the tank would be equal to 39.47 m^3 or 43.52 tons of water. So if you only have 39.47 m^3 in the dock then placed the pontoon in, it would then float to the prescribed clearances of 0.1 m from bottom and sides without spilling any water.
Now I would say your lectures volume is slightly off, but I could be wrong.
This is calculating the volume of water along the 40m long sides. Basically it is the area of the rectangles of water times the 40 m length
A1= 4*0.1 (area under pontoon)
A2= 2.934*0.1 (sidewall area to left)
A3= 2.934*0.1 (sidewall area to right)
pull out the 0.1 and simplify
What I believe is missing is the 0.1 clearance in the front and the back which has not been accounted for.
So the volume remaining in the tank should really be
39.47+2.46 = 41.93 m^3
Based on the calculations provided.
An important note here is that the height of the pontoon is not given. But the question states that it will float, so you have to assume the height is greater than 2.834m. Otherwise the pontoon is too dense and a denser medium than water would be required to make it float.
My version of the calculations. Density of water may be slightly different.