Ionic product of water at $120°C$ is ${10}^{-12}$. Concentration of hydroxyl ion in an aqueous solution of ${10}^{-6}M$ NaOH at this temperature is

$$\begin{gathered} H_{2}O \stackrel{ }{\rightleftharpoons } H^{+}+ O{H}^{-} \\ Initially 0 {10}^{-6} \end{gathered}$$
Change  +x   +x
At eqbm.  x  ${10}^{-6}+x$
$x({10}^{-6}+x)={10}^{-12}$
You may solve the quadratic equation and get the answer or you may get the answer logically.
In the absence of NaOH, water gives 10-6M hydroxyl ion. Presence of NaOH suppresses the ionization of water and hence the concentration of hydroxyl ion coming from it will be definitely less than 10-6 M. 
${10}^{-6}<{\left[O{H}^{-}\right]}_{total}<2\times {10}^{-6}M$