The decreasing order of basicity is:
(1) $C{{H}_{3}}CON{{H}_{2}}$
(2) $C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}$
(3) $Ph-C{{H}_{2}}CON{{H}_{2}}$

(A) 1 > 2 > 3
(B) 2 > 1 > 3
(C) 3 > 2 > 1
(D) None of the above

Hint: It should be kept in mind that two of the given compounds have an amide functional group and one compound has an amine functional group. The basicity of a compound means that a compound can easily donate its lone pair of electrons to acid. The greater the tendency to donate a lone pair, the greater the basicity of the compound.

Complete Step by Step Solution:
Amines are stronger bases as compared to amides. The reason for this is that the nature of the alkyl group (R) is electron releasing, so it shifts the electron density towards nitrogen and thus makes available the unshared electron pair for the proton of acid. Thus, amines are highly basic. But there is a presence of carbonyl groups in amides, which is electron withdrawing in nature. It pulls the electron density away from the nitrogen atom, which delocalizes the carbonyl group through resonance. This makes the lone pair of nitrogen unavailable for the proton of acid. Hence, compound 2 will be more basic than both compounds 1 and 3.

Now, Both compounds 1 and 3 are amides. The comparative basicity of these two amides can be explained as that in $Ph-C{{H}_{2}}CON{{H}_{2}}$ , the $Ph-C{{H}_{2}}$ group is electron-withdrawing as the resonance in phenyl will stabilise the electron pair and thus lower its basicity. So, compound 3 has a basicity lower than compound 1.
Correct Option: (B) 2 > 1 > 3.

Note: It should always be remembered that amines have always higher basicity than amides. Also, the electron withdrawing group decreases the basicity of amides, whereas the electron-releasing group increases the basicity.