Volume required to dilute solution for a pH change
Since the amount of substance stays equal, you can calculate the volume of your final solution $V_\mathrm{end}$, form your start concentration $c_\mathrm{start}(\ce{OH-})$ and your initial volume $V_\mathrm{start}$ and your target concentration $c_\mathrm{end}(\ce{OH-})$. Your working equation should be:

\begin{align}
c_\mathrm{start}(\ce{OH-})\cdot V_\mathrm{start} 
&amp;= c_\mathrm{end}(\ce{OH-})\cdot V_\mathrm{end} \\
V_\mathrm{end} &amp;= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start}
\end{align}

You can work out the start concentration for the initial pH and the target concentration from your final pH. Given (at $T=298.15~\mathrm{K}$, $p=1~\mathrm{atm}$):

\begin{align}
\ce{pH}_\mathrm{start} &amp;= 13 \\
c_\mathrm{start}(\ce{OH-}) &amp;= 10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.1~\mathrm{mol/L} \\
V_\mathrm{start} &amp;= 0.1~\mathrm{L} \\
\ce{pH}_\mathrm{end} &amp;=11 \\
c_\mathrm{end}(\ce{OH-}) &amp;=10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.001~\mathrm{mol/L}
\end{align}

Hence

$$
V_\mathrm{end}= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start} = 10~\mathrm{L}
$$

and therefore

$$
\Delta V = V_\mathrm{end}-V_\mathrm{start} = 9.9~\mathrm{L}= 9900~\mathrm{mL}.
$$