= Solution
For equal mean <mass densities>, $M_a/M_E=(r/R_E)^3$. Thus the asteroid's surface gravitational <acceleration> is $g_a=g_Er/R_E$ and its <escape speed> is
$$
v_{\rm esc}^2=\frac{2GM_a}{r}=\frac{2g_Er^2}{R_E},\qquad
\boxed{r_{\max}=v_J\sqrt{\frac{R_E}{2g_E}}.}
$$
An explicit estimate of jumping ability is an <Earth> jump height $H=0.50\,\mathrm m$, giving $v_J=\sqrt{2g_EH}\simeq3.1\,\mathrm{m\,s^{-1}}$. Consequently
$$
\boxed{r_{\max}=\sqrt{R_EH}\simeq1.8\,\mathrm{km}.}
$$
This is an order-of-magnitude estimate, scaling with assumed take-off <speed>; no unique human jump <speed> is supplied. A spherical nonrotating asteroid and negligible atmosphere are assumed.
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