Assume that algae need carbon, nitrogen and phosphorus in the atomic ratio 106:16:1

Limiting reagent questions demand that you compare compounds or components on a mole to mole basis. In order to do that you need to take these masses and convert them into moles.

Take one liter of the water.

20 mg C
0.80 mg N
0.16 mg P

divide by their respective molar masses (C = 12.01 g/mol, N = 14.01 g/mol, P = 30.97 g/mol)

C = 1.67 x 10^-3 moles/L
N = 5.71 x 10^-5 moles/L
P = 5.17 x 10^-6 moles/L

Now that we have them on a relative molar scale in terms of their absolute numbers, we have to use the atomic ratios to further compare.

We now take each of the moles/L and divide by their respective atomic ratio number to normalize each value for direct comparison.

C = 1.67 x 10^-3/106 = 1.57 x 10^-5
N = 5.71 x 10^-5/16 = 3.57x 10^-6
P = 5.17 x 10^-6/1 = 5.17 x 10^-6

Now the above numbers are available to be directly compared on a one on one basis...

As can be seen from the above numbers, you can observe that nitrogen is the limiting reagent (smallest number) and that limiting phosphorous pollution will not reduce the algae.