I'm curious how much $\ce{D2O}$ (heavy water) is in Vienna Standard Mean Ocean Water. I am interested in $\ce{D2O}$ molecules and not $\ce{HDO}$/$\ce{D+}$/ $\ce{DH2O+}$/$\ce{D2HO+}$/$\ce{D3O+}$ or any other deuterium species or ions. Vienna Standard Mean Ocean Water is pure water extracted from some ocean with a specific isotopic composition. Sweet water has slightly different isotopic composition.
A suitable unit for the result could be mol/kg or a mol fraction.
I'm not sure if temperature or pressure make a difference. I suggest $\pu{20 ^\circ C}$ (or any other temperature, please specify, if possible) and $\pu{101325 Pa}$, but other conditions are ok, too. Liquid water makes most sense, I guess.
Allright, I took some effort as suggested and used my brain ...
2H/1H = 155.76 ± 0.1 ppm (a ratio of 1 part per approximately 6420 parts)1
So if an HDO autoprotolyzed ...
HDO --> D+ + OH- or HDO --> H+ + OD-
and then D+ or OD- reformed water with OH- or H+, then it would have a 1 in 6420 chance of forming D2O, and so D2O would be 1 in 6420^2=41,216,400 parts.
Sounds good?
1 https://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water