The most basic hydroxide among the following is

Basicity is a measure of the ease at which an atom will lose electron. For the lanthanides, the basicity decreases as the atomic number increases.

The order of the basicity for the lanthanides are${\mathrm{La}}^{3+} > {\mathrm{Ce}}^{3+} > {\mathrm{Pr}}^{3+} > {\mathrm{Nd}}^{3+} > {\mathrm{Pm}}^{3+} > {\mathrm{Sm}}^{3+} > {\mathrm{Eu}}^{3+} > {\mathrm{Gd}}^{3+} > {\mathrm{Tb}}^{3+} > {\mathrm{Dy}}^{3+} > {\mathrm{Ho}}^{3+} > {\mathrm{Er}}^{3+} > {\mathrm{Tm}}^{3+} > {\mathrm{Yb}}^{3+} > {\mathrm{Lu}}^{3+}$

and also ${\mathrm{Ce}}^{+4}$ is less stable than ${\mathrm{Ce}}^{+3}$and thus it will readily accept ${\mathrm{e}}^{-}$  convert into ${\mathrm{Ce}}^{+3}$ and so it will less basic than ${\mathrm{Ce}}^{+3}$.

Hence, from all above observations it can be concluded that $\mathrm{Ce}{\left(\mathrm{OH}\right)}_3$is most basic among all given options.