Degree of dissociation of acetic acid at infinite dilution

Afew drops of acetic acid are added to a large excess of water at $$\pu{25 ^\circ C}$$. In the limit of infinite dilution, what percentage of acid remains unionised? The $$K_\mathrm{a}$$ of acetic acid is $$1.85\times10^{-5}$$.

The answer is supposed to be 0.54%, but I'm not sure how to reach this value. I know that given the pH we can use the definition of $$K_\mathrm{a}$$ to calculate the ratio of $$\ce{CH3COOH}$$ to $$\ce{CH3COO-}$$, but how do I find the pH?

\begin{align} \frac{(10^{-7})[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 1.85 \times 10^{-5} \\ [\ce{CH3COO-}] &= 185 [\ce{CH3COOH}] \end{align}
The fraction of unionised acid, $$x$$, is therefore given by
\begin{align} x &= \frac{[\ce{CH3COOH}]}{[\ce{CH3COOH}] + [\ce{CH3COO-}]} \\ &= \frac{1}{1+185} \\ &= 0.54\%. \end{align}