= Solution
For the unresolved <Lyman-alpha line>, the lower $1s$ and combined upper $2p$ <state degeneracies> are $g_1=2$ and $g_2=6$, so the <Einstein coefficients> give
$$
B_{12}=3\frac{A_{21}c^3}{8\pi h\nu_{12}^3}=2.03\times10^{20}\ \mathrm{m^3\,J^{-1}\,s^{-2}}.
$$
The measured <equivalent width> $W=7.0\times10^{-12}\ \mathrm m$ then implies
$$
N_{\rm HI}=\frac{8\pi W\nu_{12}^4}{3A_{21}c^3}
=1.29\times10^{17}\ \mathrm{m^{-2}}=1.29\times10^{13}\ \mathrm{cm^{-2}}.
$$
Assume a central transverse <line of sight> through a uniform filament, so $\ell=d=0.5\ \mathrm{Mpc}=1.54\times10^{22}\ \mathrm m$. Nearly all neutral <hydrogen> is in its ground state at this <temperature>, hence
$$
\boxed{n_{\rm HI}=N_{\rm HI}/d=8.3\times10^{-6}\ \mathrm{m^{-3}}=8.3\times10^{-12}\ \mathrm{cm^{-3}}.}
$$
The cylinder's length $L$ determines its total volume but does not enter a transverse <column density>. Geometry must be specified: an axial <line of sight> has $\ell=L$ and gives $n_{\rm HI}=4.2\times10^{-7}\ \mathrm{m^{-3}}$ for the same measured <equivalent width>. Intermediate chords give intermediate path-dependent estimates. The numbers above treat $W$ as a rest-frame <equivalent width>; an observed value at <redshift> $z$ must first be divided by $1+z$.
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