= Solution
Conservation of the comoving galaxy population gives $n(z)=n_0a^{-3}=n_0(1+z)^3$. Holding the proper galaxy radius fixed, integrate <4/a>:
$$
\tau(z_s)=\frac{c\pi r_G^2n_0}{H_0}\int_0^{z_s}(1+z)^{1/2}dz
=\boxed{\frac{2c\pi r_G^2n_0}{3H_0}\left[(1+z_s)^{3/2}-1\right].}
$$
With $r_G=0.015\ \mathrm{Mpc}$, $n_0=0.01\ \mathrm{Mpc^{-3}}$, and $c/H_0=4282.75\ \mathrm{Mpc}$, set $C=c\pi r_G^2n_0/H_0=0.030273$. Consequently
$$
\boxed{\tau(1)=0.0369013.}
$$
For random uncorrelated intersections, the zero-intersection probability is $e^{-\tau}$ from the <Poisson distribution>, so the fraction of the sky covered by foreground galaxies \i[out to] $z=1$ is
$$
\boxed{f=1-e^{-\tau(1)}=0.03623\simeq3.6\%.}
$$
Ignoring overlaps gives $f\simeq\tau\simeq3.7\%$. A literal shell of zero thickness exactly at $z=1$ has zero coverage; the integrated foreground interpretation is needed. Clustering changes the exact overlap probability, although the mean <optical depth> above still follows from the mean <number density>.
Finally,
$$
\boxed{\tau=1\quad\Longrightarrow\quad z_s=\left(1+\frac3{2C}\right)^{2/3}-1=12.67.}
$$
At this <optical depth>, the Poisson sky coverage is $1-e^{-1}\simeq63\%$, not $100\%$. The large inferred <cosmological redshift> extrapolates conserved galaxy counts and fixed radii far beyond their plausible physical range; it is the answer within the prescribed toy model.
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