Calculating the pH of a dilluted aqueous KOH solution
Yes, that's right, the concentration of $\ce{{}^{-}OH}$ from dissociation of water will be much greater than the amount added from $0.01\:\mathrm{nM}~\ce{KOH}$.  

$0.01\:\mathrm{nM}~\ce{KOH}$ will not significantly the affect $\mathrm{p}\ce{H}$ of water.