The increasing order of pK$$_b$$ for the following compounds will be:

The three given amines are:

$$\text{(A)} : C_6H_5NH_2 \;(\text{aniline})$$
$$\text{(B)} : (CH_3)_2NH \;(\text{dimethylamine})$$
$$\text{(C)} : CH_3NH_2 \;(\text{methylamine})$$

Step 1: Recall the relation between basicity, $$K_b$$ and $$pK_b$$.
For any base, $$K_b = \dfrac{[\text{BH}^+][OH^-]}{[\text{B}]}$$ and $$pK_b = -\log_{10}K_b$$.
Higher basic strength ⇒ larger $$K_b$$ ⇒ smaller $$pK_b$$.
Therefore, arranging compounds in decreasing basicity is equivalent to arranging them in increasing $$pK_b$$.

Step 2: Compare the basic strengths.

(i) Dimethylamine, $$(CH_3)_2NH$$, has two $$+I$$ methyl groups that push electron density toward the nitrogen. This stabilises the conjugate acid $$(CH_3)_2NH_2^{+}$$ and makes the lone pair more available for protonation. Hence it is strongly basic.

(ii) Methylamine, $$CH_3NH_2$$, possesses only one $$+I$$ methyl group, so the electron-donating effect is smaller. Its basicity is therefore lower than that of dimethylamine but higher than that of aniline.

(iii) In aniline, $$C_6H_5NH_2$$, the lone pair on nitrogen is delocalised into the aromatic ring through resonance:
$$C_6H_5NH_2 \;\longleftrightarrow\; C_6H_5\!\!-\!\!N^{+}H_2 \; \text{(with } \!\!=\!\! \text{ in the ring)}$$
Because the lone pair participates in resonance, it is less available to accept a proton, so aniline is the weakest base of the three.

Thus the order of basic strength (largest to smallest) is
$$ (CH_3)_2NH \; \gt \; CH_3NH_2 \; \gt \; C_6H_5NH_2 $$ and the corresponding order of $$pK_b$$ (smallest to largest) is

$$ (CH_3)_2NH \; \lt \; CH_3NH_2 \; \lt \; C_6H_5NH_2 $$

Step 3: Map the order to the question’s labels.
$$\text{(B)} \lt \text{(C)} \lt \text{(A)}$$

Hence, the increasing order of $$pK_b$$ is (B) < (C) < (A).

Option A which is: (B) < (C) < (A)