= Solution
Set $x=0.1x_0$:
$$
\boxed{t_{90,A}\simeq\frac{\ln10}{k(10^{-2}\,\mathrm M)}},
\qquad
\boxed{t_{90,B}=\frac9{k(10^{-4}\,\mathrm M)}},
$$
$$
\boxed{t_{90,C}=\frac{\ln10}{k(10^{-4}\,\mathrm M)}}.
$$
The corresponding numerical <concentration> factors are approximately $230.3$, $90000$ and $23026\,\mathrm{M^{-1}}$, each divided by $k$. Thus A is about 100 times faster than C at this conversion, while B takes about $9/\ln10=3.91$ times as long as C because its <hydroxide> is depleted.
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