The compound that will react most readily with NaOH to form methanol is

Compound ${\left({\mathrm{CH}}_3\right)}_4{\mathrm{N}}^{+}{\mathrm{I}}^{-}$ is most reactive due to (i) better leaving group, $I^{-}$ and (ii) due to the fact that the methyl group, with positive N, is not more electron deficient. Hence this group is more reactive towards nucleophile, $O{H}^{-}$