= Solution
Use $R$ for the physical projected radius, or impact parameter, in the lens plane. Although the source calls it angular, a physical length is required in the mass and deflection equations; its relation to an angle is $R=D_L|\theta|$. Integrate along physical line-of-sight coordinate $z$:
$$
\Sigma(R)=\int_{-\infty}^{\infty}\frac{A\,dz}{R^2+z^2}
=\frac{\pi A}{R}.
$$
Consequently the projected cylindrical mass inside the image radius is
$$
M_{\mathrm{2D}}(<R)=2\pi\int_0^R R'\Sigma(R')\,dR'=2\pi^2AR.
$$
Substituting <5/a> gives
$$
\boxed{\Sigma(R)=\frac{\sigma_{\mathrm{DM}}^2}{2GR},\qquad
M_{\mathrm{2D}}(<R)=\frac{\pi\sigma_{\mathrm{DM}}^2}{G}R.}
$$
If an angular radius $\vartheta$ is used instead, write $\Sigma(D_L\vartheta)=\sigma_{\mathrm{DM}}^2/(2GD_L\vartheta)$ and $M_{\mathrm{2D}}(<\vartheta)=\pi\sigma_{\mathrm{DM}}^2D_L\vartheta/G$. The projected mass differs from the spherical mass $2\sigma_{\mathrm{DM}}^2R/G$; material at three-dimensional radii larger than $R$ still projects into the cylinder.
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