The pH of 0.01 M NaOH (aq) solution will be

$\begin{array}{l}\left[{\mathrm{OH}}^{-}\right]=\left[\mathrm{NaOH}\right]=0.01\mathrm{M}={10}^{-2}\mathrm{M}\\ \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log \left({10}^{-2}\right)\\ =2\\ \mathrm{pH}=14-\mathrm{pOH}=12\end{array}$