Step 1: Link \(pK_b\) to base strength.
A smaller \(pK_b\) value means a stronger base. So we must find the amine in which the nitrogen lone pair is most available to accept a proton.
Step 2: Recall the effect of alkyl groups on aromatic amines.
In aromatic amines, replacing the N-H hydrogens with electron-donating alkyl groups (\(-CH_3\)) pushes electron density onto nitrogen, making it more basic.
Step 3: Compare the given amines.
Aniline \(C_6H_5NH_2\) has no extra alkyl groups, N-methylaniline \(C_6H_5NH(CH_3)\) has one, and N,N-dimethylaniline \(C_6H_5N(CH_3)_2\) has two.
Step 4: Order the basicity.
More methyl groups on nitrogen means more electron density and stronger base, so \(C_6H_5N(CH_3)_2 > C_6H_5NH(CH_3) > C_6H_5NH_2\) in basicity.
Step 5: Conclude.
The strongest base, and hence the lowest \(pK_b\), is N,N-dimethylaniline \(C_6H_5N(CH_3)_2\).
\[ \boxed{C_6H_5-N(CH_3)_2} \]