= Solution
For <circular orbits in an exponential central potential>, a constant positive radius requires $V_{\rm eff}'(r)=0$, or
$$
l^2=r^3e^{-r}.
$$
Let $h(r)=r^3e^{-r}$. Its derivative is $h'(r)=r^2e^{-r}(3-r)$: it rises from zero to $h(3)=27/e^3$, then falls back to zero. Hence
$$
\boxed{0<l^2<27/e^3\quad\Longrightarrow\quad
0<r_1<3<r_2.}
$$
The strict lower bound matters: the printed upper bound also allows $l=0$, but then there is no finite-radius <circular orbit>. Nonzero <angular momentum> is an implicit hypothesis of the two-orbit claim.
At a <circular orbit>, use $l^2=r^3e^{-r}$ to obtain
$$
V_{\rm eff}''(r)=-e^{-r}+\frac{3l^2}{r^4}
=e^{-r}\left(\frac3r-1\right).
$$
A small radial displacement $\eta$ satisfies $\ddot\eta=-V_{\rm eff}''(r)\eta$. Therefore \b[the inner <circular orbit> is radially stable at fixed <angular momentum>, while the outer orbit is unstable]. The inner perturbation oscillates with squared frequency $e^{-r_1}(3/r_1-1)$; the outer perturbation has an exponentially growing solution. At the limiting value $l^2=27/e^3$, the radii coalesce at $r=3$ and the linear restoring coefficient vanishes.
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