Order of magnitude of ions in water My book says water has approximately $\pu{3.35E28 molecules/m^3}$  and that only a small part of water molecules are dissociated to $\ce{H+}$ and $\ce{OH-}$ ions and that the approximate amount of ions of each sign is $\pu{6E19 ions/m^3}.$  

I am trying to prove the last part. I thought I could used the ionic product of water

$$[\ce{H+}][\ce{OH-}] = \pu{1E-14},$$

so that in neutral conditions

$$[\ce{H+}] = \pu{1E-7 ions/L} = \pu{1E-4 ions/m^3}.$$ 

What am I doing wrong or how should I proceed?
You are right: 1 cubic meter of water contains $\pu{1E-4}$ moles of $\ce{H+}$ ions. But one mole is $\pu{6E23}$ atoms or ions.

So, $\pu{1E-4 mol}$ contains $\pu{1E-4}\times\pu{6E23} = \pu{6E19 ions}.$
Here we are.