The percentage of pyridine $\left(C_5{H}_{5}N\right)$ that forms pyridinium ion $\left(C_5{H}_5{N}^{+}H\right)$ in a 0.10 M aqueous pyridine solution ($K_b$ for $C_5{H}_{5}N=1.7\times {10}^{-9}$) is

The percentage of pyridine can be equal o the percentage of dissociation of pyridinium ion and pyridine solution as shown below :

As pyridinium is a weak base, so degree of dissociation is given as
$\alpha =\sqrt{\frac{K_b}{C}}=\sqrt{\frac{1.7\times {10}^{-9}}{0.10}}=\sqrt{1.7\times {10}^{-8}}=1.3\times {10}^{-4}$
or, percentage of dissociation $=\left(\alpha \times 100\right)\%$
$=\left(1.3\times {10}^{-4}\right)\times 100$
$=0.013\%$