I'm no chemist. With my limited learning on this, I come up with about 4.2% Lugol's solution. That's based on several assumptions: you added exactly 44 g iodine and 88 g potassium iodide, and that the 32 ounces of water had a mass of approximately 907 g. 44g+88g+907g=1039g total solution. 44 divided by 1039 = .042 = 4.2% iodine in the solution, which is 4.2% Lugol's.
To figure mg per drop, I'd round down to 4%. If you're using a dropper size known to deliver 6.25 mg iodine/iodide per drop of 5% Lugol's, you might expect per drop from this solution 2 mg iodine/3 mg iodide = 5 mg iodine/iodide.
You could argue for the other 0.2% (referring to 4.2%), but that would add only 0.25 mg per drop. Given the unknowns, I don't think it's worth believing either one of us is that accurate.