= Solution
For the <binding energy of a truncated singular isothermal sphere>, build the mass distribution from concentric shells. Its <enclosed mass> and mass element are
$$
M(r)=\int_0^r4\pi s^2\rho(s)\,\mathrm ds
=\frac{2k}{G}r,\qquad
\mathrm dM=\frac{2k}{G}\,\mathrm dr,
\qquad M_0=\frac{2kr_0}{G}.
$$
The interaction of a newly added shell with the interior contributes $-GM(r)\,\mathrm dM/r$. Counting each pair once,
$$
W=-\int_0^{r_0}\frac{GM(r)}r\,\mathrm dM(r)
=-\frac{4k^2}{G}\int_0^{r_0}\mathrm dr
=-\frac{4k^2r_0}{G}.
$$
Eliminating $k$ gives
$$
\boxed{W=-\frac{GM_0^2}{r_0}.}
$$
The central density is singular, but the mass and this energy integral are finite. The result assumes the stated truncation and no additional gravitating component.
The <gravitational potential of a truncated singular isothermal sphere> is a separate quantity from this shell-assembly energy allocation. With zero potential at infinity, both interior and exterior shells contribute:
$$
\Phi_{\rm grav}(r)
=-G\left[\frac{M(r)}r+\int_r^{r_0}\frac{\mathrm dM(s)}s\right]
=-2k\left[1+\ln\frac{r_0}r\right],\quad0<r\leq r_0.
$$
It matches $-GM_0/r_0=-2k$ at the edge. The equivalent total energy is $W=\tfrac12\int\rho\Phi_{\rm grav}\,\mathrm dV$; the half prevents double counting. By contrast, $\int\rho[-\Phi_{\rm grav}]\,\mathrm dV$ counts the escape work of test parcels in a fixed potential and equals $-2W$. Confusing these quantities would change a local feedback threshold.
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