How to calculate the number of protons in a sphere given the volume and the pH?
As pointed out in the comments, the concept of pH as a thermodynamic property of nanodroplets does not quite make sense unless the nanodroplets are in equilibrium (stable). If they are in equilibrium you can still apply the equation 
$$\textrm{pH}=-\log_{10}(a_{\ce{H+}})$$ but now you have to be aware that the activity will be influenced by surface (and other) effects and the approximation $$\textrm{pH}=-\log_{10}(c_{\ce{H+}})$$ may not hold. But assume it does. Then as you point out, the average number of hydronium ions per nanodroplet is 

$$\begin{align} n_{\ce{H+}} &amp;=\pu{10^{-6} M}\cdot(\frac{4}{3}\pi\times\pu{20^3 nm^3}\times\pu{10^{-27} {(m/nm)^3}}\times\pu{10^3 L/m^3})\times \pu{6.022\times10^{23} mol^{-1}}\\&amp;=0.020 \end{align}$$ 

Which is ok. That just means that on average only 1 out of 50 nanodroplets contains a solvated proton. Note by the way that one nanodroplet also contains about a million water molecules.

This is a real "rabbit hole" of a question, given all that is not specified, which is why I used sleight-of-hand and wrote that I am ignoring surface effects and assuming the particles are at equilibrium. It doesn't address how to prepare nanodroplets with the desired average proton concentration or what equilibrium means (are nanodroplets exchanging matter or are they isolated?). As noted in comments, composition (buffering molecules if any, salts and other ions including hydroxyl), droplet charge etc, are being ignored and you can expect statistical fluctuations in concentrations between different droplets. However if I stick to the assumption of equilibrium then I can say (for what it's worth) that ~2% of droplets have ~pH 4.3, and most of the rest are at neutral pH.