The rate law for the substitution reaction of 2-bromobutane and  $O{H}^{-}$ in 75% ethanol – 25% water at $30°\mathrm{C}$ is
$\text{Rate =3.2}\times {\text{10}}^{-5}[2-bromobu\tan e]\left[O{H}^{-}\right]+2\times {10}^{-6}[2-bromobu\tan e]$ . 
Then what is the percentage of the reaction taking place by the $S{N}_2$ mechanism when $\left[O{H}^{-}\right]$ =1.0M