I'll take a crack at it. I'm going to assume an average turbidity of flood waters of 100 mg/l. I got this from looking up what a high turbidity for fish ponds is and going with the the too high value. If you are dealing with levels that figure higher than that and the fish are going to die off. (I know, there are other things that would kill off fish in a global flood, but since we still have fish, let's put the others to the side and just look at suspended particulates not killing fish)
Now, next step, let's look at what kind of volume we are talking about from those suspended solids. Let's assume an average density of 2.5 g/cm3 assuming that limestone is going to be typical of such sediment. Now, to play with units until we get depth of sediment per depth of water
100 mg/l = .1 g/l = .0001 g sediment / cm3 water
Divide that by 2.5 g sediment / cm3 sediment
.0001 / 2.5 = 0.00004 cm3 sediment / cm3 water
but we want depth to depth, so let's take the cube root of that and we get
0.0342 units of sediment per unit of water.
Now, let's call the flood depth 9000 m above sea level. That means at sea level, we should find about 300 m of sediment with less thickness higher above sea level, and more below. At the Mariana trench, we would be looking at about 19000 m total depth, or about 650 m of sediment.