 Solution

In the local, nonrelativistic Hubble law approximation, $c{z}_{\mathrm{tot}}\simeq H_{0}r+v_p$. Ignoring the peculiar velocity gives an inferred distance $r_{\mathrm{est}}=r+v_p/H_0$, soThe requested accuracy therefore requiresAt 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 $r=c{z}_{\mathrm{cosm}}/H_0$ as exact, the positive-velocity case gives $|\Delta r|/r=(1+z_{\mathrm{cosm}})q/z_{\mathrm{cosm}}$, with $q=\sqrt{(1+600/c)/(1-600/c)}-1$. It requires $r>(c/H_0)q/(0.05-q)\simeq 125\mathrm{Mpc}$. That small refinement is conditional on the distance-redshift prescription; the simple $120\mathrm{Mpc}$ answer uses the local approximation consistently.