Step 1: Find the hydrogen ion concentration:
$[\text{H}^+] = \sqrt{K_aC} = \sqrt{1.96\times10^{-8}\times10^{-2}} = \sqrt{1.96\times10^{-10}} = 1.4\times10^{-5}$ M.
Step 2: Divide by the initial concentration:
$\alpha = [\text{H}^+]/C = 1.4\times10^{-5}/10^{-2} = 1.4\times10^{-3}$.
Step 3: Convert to percentage:
$1.4\times10^{-3}\times100 = 0.14\%$.
Step 4: Check:
Since $\alpha \ll 1$ the weak-acid approximation holds. The choice 1.40 % is exactly ten times larger, which would be the result if $K_a$ were $1.96\times10^{-6}$.
Final Answer:
Percentage dissociation $= 0.14\%$, option (B).
\[ \boxed{0.14\,\% \text{ (B)}} \]