= Solution
For the <singular isothermal sphere>, integrate its <projected surface mass density> inside a circle of proper radius $\xi$:
$$
M_{\rm proj}(<\xi)=2\pi\int_0^\xi\frac{\sigma^2}{2G\xi'}\xi'\,d\xi'
=\frac{\pi\sigma^2\xi}{G}.
$$
With $\xi=D_d\theta$ this gives the reduced deflection magnitude
$$
\frac{D_{ds}}{D_sD_d}\frac{4GM_{\rm proj}(<D_d\theta)}{c^2\theta}
=4\pi\frac{\sigma^2}{c^2}\frac{D_{ds}}{D_s},
$$
independent of $\theta$. Hence the <singular isothermal sphere lens> has
$$
\boxed{\theta_E=4\pi\left(\frac\sigma c\right)^2\frac{D_{ds}}{D_s}.}
$$
The spherical three-dimensional mass $2\sigma^2r/G$ would give the wrong coefficient here; the <thin gravitational lens equation> needs the projected cylindrical mass.
Using the same $\theta_E=3\times10^{-4}$ and $D_{ds}/D_s=0.5$, the one-dimensional <velocity dispersion> is
$$
\sigma=c\sqrt{\frac{\theta_E}{4\pi(D_{ds}/D_s)}}
=3\times10^8\sqrt{\frac{3\times10^{-4}}{2\pi}}\,\mathrm{m\,s^{-1}}
\simeq2.1\times10^6\,\mathrm{m\,s^{-1}}.
$$
Thus \b[$\sigma\simeq2100\,\mathrm{km\,s^{-1}}$] within the idealized <SIS> model.
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