$$ \begin{aligned} &{\left[\mathrm{OH}^{-}\right]=0.001=10^{-3} \mathrm{M}} \\\\ &{\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}} \\\\ &\quad\left[\mathrm{H}^{+}\right]=10^{-11} \\\\ &\begin{aligned} \mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\\\ =&-\log \left(10^{-11}\right) \\\\ \mathrm{pH} &=11 \end{aligned} \end{aligned} $$