= Solution
A complete <Einstein ring> requires an axisymmetric <gravitational lens> and exact alignment of the source with the lens centre and observer: $\beta=0$. A finite aligned circular source gives a ring of finite width. The background source must lie behind the lens, so the lensing distance factor is positive.
With $\beta=0$ the <thin gravitational lens equation> gives
$$
\boxed{\theta_E^2=\frac{4G}{c^2}\frac{D_{ds}}{D_sD_d}\,M(<D_d\theta_E).}
$$
The <Einstein radius> here is angular, and the mass is projected inside the corresponding aperture. For an extended mass distribution the equation is implicit because the enclosed mass depends on $\theta_E$.
Using the supplied approximations, $D_{ds}/(D_sD_d)=1/(3\times10^{25}\,\mathrm m)$ and $\theta_E=3\times10^{-4}$, the inferred mass is
$$
M=\frac{\theta_E^2c^2}{4G}\frac{D_sD_d}{D_{ds}}
=\frac{(3\times10^{-4})^2(3\times10^8)^2(3\times10^{25})}{4(7\times10^{-11})}\,\mathrm{kg}
\simeq8.7\times10^{44}\,\mathrm{kg}.
$$
Thus
$$
\boxed{M(<\theta_E)\simeq4.3\times10^{14}M_\odot.}
$$
This measures the projected aperture mass without requiring the <singular isothermal sphere> model used subsequently.
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