= Solution
In the local, nonrelativistic <Hubble law> approximation, $cz_{\rm tot}\simeq H_0r+v_p$. Ignoring the <peculiar velocity> gives an inferred distance $r_{\rm est}=r+v_p/H_0$, so
$$
\frac{|r_{\rm est}-r|}{r}\leq\frac{600}{100r}
=\frac{6\,\mathrm{Mpc}}{r}.
$$
The requested accuracy therefore requires
$$
\boxed{r>120\,\mathrm{Mpc},\qquad r_{\min}\simeq120\,\mathrm{Mpc}.}
$$
At exactly $120\,\mathrm{Mpc}$ the worst-case error is $5\%$, so “better than” requires a strictly larger distance. This is the conventional leading-order estimate. If one retains the exact Doppler factor and the product term while treating $r=cz_{\rm cosm}/H_0$ as exact, the positive-velocity case gives $|\Delta r|/r=(1+z_{\rm cosm})q/z_{\rm cosm}$, with $q=\sqrt{(1+600/c)/(1-600/c)}-1$. It requires $r>(c/H_0)q/(0.05-q)\simeq125\,\mathrm{Mpc}$. That small refinement is conditional on the distance-redshift prescription; the simple $120\,\mathrm{Mpc}$ answer uses the local approximation consistently.
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