Step 1: Calculate the hydrogen ion concentration from the given pH.

$$pH = 3.44$$

$[H^+] = \text{antilog}(-pH) = \text{antilog}(-3.44) = 3.63 \times 10^{-4}$ M

Step 2: Write the hydrolysis reaction and the relation for $[H^+]$. Pyridinium hydrochloride ($C_5H_5NH^+Cl^-$) is a salt of a weak base and a strong acid. The cation hydrolyzes as:

$$C_5H_5NH^+ + H_2O \rightleftharpoons C_5H_5N + H_3O^+$$

For this salt, $[H^+] = \sqrt{\frac{K_w \times C}{K_b}}$

Step 3: Rearrange the formula to solve for $K_b$.

$$K_b = \frac{K_w \times C}{[H^+]^2}$$

Step 4: Substitute the known values into the equation. Given: $C = 0.02$ M, $K_w = 1 \times 10^{-14}$, $[H^+] = 3.63 \times 10^{-4}$ M

$$K_b = \frac{1 \times 10^{-14} \times 0.02}{(3.63 \times 10^{-4})^2}$$

Step 5: Calculate the final value of $K_b$.

$$K_b = \frac{2 \times 10^{-16}}{13.17 \times 10^{-8}} = 1.52 \times 10^{-9}$$