= Solution
Take $d$ as the centre-to-centre separation and $\epsilon=\epsilon_0\epsilon_r$. Use the ideal lossless two-conductor <transmission line> model, with nonmagnetic surroundings and negligible internal-wire <inductance>. Opposite <line charges> $+\lambda_q$ and $-\lambda_q$ approximate the cylindrical conductors when $d\gg a$. Integrating their <electric fields>, or adding their logarithmic <electric potentials>, gives the potential difference
$$
V\simeq\frac{\lambda_q}{\pi\epsilon}\ln\frac da,
\qquad C'=\frac{\lambda_q}{V}\simeq\frac{\pi\epsilon}{\ln(d/a)}.
$$
This is the <two-wire capacitance at large separation>. For opposite <electric currents> $+I$ and $-I$, the <magnetic fields> between the wires add. The linked <magnetic flux> per unit length is approximately
$$
\Phi'\simeq\frac{\mu_0I}{2\pi}\int_a^{d-a}\left(\frac1x+\frac1{d-x}\right)dx
=\frac{\mu_0I}{\pi}\ln\frac{d-a}{a}
\simeq\frac{\mu_0I}{\pi}\ln\frac da.
$$
Thus the <external inductance of a thin two-wire line> is $L'\simeq\mu_0\ln(d/a)/\pi$. The lossless <telegrapher's equations> give a forward <wave> with $V/I=\sqrt{L'/C'}$, so its <characteristic impedance> is
$$
\boxed{Z_0\simeq\frac1\pi\sqrt{\frac{\mu_0}{\epsilon_0\epsilon_r}}\ln\frac da
\simeq\frac{120\,\Omega}{\sqrt{\epsilon_r}}\ln\frac da.}
$$
The product $L'C'=\mu_0\epsilon$ also gives the propagation speed $c/\sqrt{\epsilon_r}$, an independent consistency check. The exact ideal external-field geometry replaces $\ln(d/a)$ by $\operatorname{arcosh}(d/(2a))$. If the wire separation were specified between surfaces, its centre separation would be $d+2a$, with the same leading large-separation result. Finite <electrical conductivity> and internal-wire <magnetic energy> require a more detailed line model.
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