Explanation
$$1 \ ppm = 1 \ mg/L = 10^{-3} g/L$$
Water sample contains $$1 \ ppm$$ urea concentration.
$$\therefore \ 10^{-3} \ g$$ of urea in $$ 1 \ L$$
$$x \ g$$ urea in $$0.05 \times 10^{-3} \ L$$
$$x= 0.05 \times 10^{-6} \ g$$
$$60 \ g$$ urea $$=6.023 \times 10^{23} \ molecules$$
$$0.05 \times 10^{-6} \ g = n \ molecules$$
$$n = \cfrac {6.023 \times 10^{23} \times 0.05 \times 10^{-6}}{60}$$
$$=0.005 \times 10^{17}$$
$$=5 \times 10^{14} \ molecules$$
We know that
$$K_a = \cfrac {K_w}{K_b}= \cfrac {10^{-14}}{1.7 \times 10^{-9}}= 5.88 \times 10^{-6}$$
$$C_5H_5N+ H_2O \rightleftharpoons C_5H_5NH + C_5H_5NH^+ + OH^-$$
Initial conc: $$0.1$$ $$0$$ $$0$$ $$0$$
Final Conc: $$0.1-x$$ $$x$$ $$x$$
$$K_a = \cfrac {x^2}{0.1-x}$$ [Since $$x <<0.1$$ it can be neglected]
$$5.88 \times 0.1 \times 10^{-6} =x^2$$
$$x=7.6 \times 10^{-4}$$
%$$ Pyridine =\cfrac {7.6 \times 10^{-4}}{0.1} \times 100 = 0.77\%$$
$$ \because \left (\cfrac {dissociated \ ions}{concentration \ of \ solution }\times 100\right)$$
Hence, the correct option is $$A$$
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