$\left(\mathrm{CH}_{3}\right)_{2} \bar {N}H>\mathrm{CH}_{3} \bar {N}H_{2}>$$\left(CH_{3}\right)_{3} \bar {N}>\mathrm{C}_{6}\ mathrm{H}_{5} \bar {N} \mathrm{H}_{2}$

As we know $\mathrm{pk}_{\mathrm{b}}=-\log \mathrm{K}_{\mathrm{b}}$

So $ (CH_3) _{2}$  $NH$ will have smallest $pK_{b}$ value. In the case of phenylamine, $N$ is attached to $s p^{2}$ hybridised carbon, hence it has highest $p K_{b}$ and least basic strength.