Not an exact method, but lets try to use the data from 10.1021/je00019a007.

Your starting binary mixture EtOH-water with x(EtOH) = 0.065 is equivalent to a mass fraction of:

w(EtOH) = 0.1509

which means that

m(EtOH)/m(water) = w(EtOH)/w(water) = 0.1777

This ratio is independent of the added Limonene. In a ternary mixture, this ratio can still be calculated as w(EtOH)/w(water).

Plotting the data from Table 2 as w(Limonene) vs. w(EtOH)/w(water) and including a point for pure water (w(EtOH)/w(water) = 0) solubility (Table 1) we get the dependence of limonene solubility on the ethanol/water ratio. Fitting the water rich phase data for range w(EtOH)/w(water) = 0 ... 1 we get approximately linear dependence:

w(limonene) = 2e-5 + 0.00165*w(EtOH)/w(water)

Using your input w(EtOH)/w(water) = 0.1777:

w(limonene) = 3.1e-4

Which is equivalent to 0.31 g of limonene per 1 kg of ethanol/water mixture.

Thanks alot Tibor! But 0.31 g limonene per 1 kg mixture seems a bit high to me. I calculated the solubility in the same way you did but with data from 10.1023/B:JOSL.0000037776.80060.b3 who seems to have more precise data for low limonene concentration. I made a log-log-plot of the data from Table III (caution! The author mixed-up the column labels) as x(limonene) vs. x(EtOH)/x(H2O), which appeared in a straight line. So I fitted the data and found the following dependence:

x(Limonene) = 1.82e-6 + 0.00457*((x(EtOH))/(x(H2O)))^(2.529)

With the given ethanol/water ratio of x(EtOH))/x(H2O) = 0,069:

x(Limonene) = 7.099e-6 bringing me to a maximum solubility of 47.6 mg/L ethanol/water mixture. Compared to the solubility in pure water (13.8 mg/L) this sounds more realistic and fits to my experiments for the Ostwald ripening process quiet well. :)

Thank you very much again for your help!