= Solution
For charge per length $\lambda_q$, <Gauss's law> gives $E_r=\lambda_q/(2\pi\epsilon_0r)$ between the conductors. Integration gives $V=\lambda_q\ln(b/a)/(2\pi\epsilon_0)$. Similarly <Ampère's law> gives $B_\phi=\mu_0I/(2\pi r)$; integrating the magnetic <energy> per length and setting it equal to $LI^2/2$ gives
$$
\boxed{C=\frac{2\pi\epsilon_0}{\ln(b/a)},\qquad L=\frac{\mu_0}{2\pi}\ln(b/a).}
$$
Here $C,L$ are per unit length. The displayed $L$ is the external TEM inductance, appropriate to the ideal surface-current/high-frequency model. Internal magnetic <energy> in conductors with appreciable current penetration adds internal inductance; it cannot be determined fully from an outer conductor with unspecified wall geometry.
The distributed series resistance and inductance, with shunt capacitance and negligible leakage, give the <telegrapher's equations>
$$
\partial_xV=-L\partial_tI-RI,\qquad\partial_xI=-C\partial_tV.
$$
Differentiate the first with respect to $x$ and substitute the second to obtain
$$
\boxed{\partial_x^2V=LC\partial_t^2V+RC\partial_tV.}
$$
For $V\propto e^{ikx-i\omega t}$ the forward attenuating branch has $k^2=\omega^2LC+i\omega RC$. When $R\ll\omega L$,
$$
\boxed{k\simeq\omega\sqrt{LC}+i\frac R2\sqrt{\frac CL},\qquad
v_{\rm ph}\simeq\frac1{\sqrt{LC}}=c,\qquad
\ell_{\rm amp}=\frac2R\sqrt{\frac LC}.}
$$
The positive imaginary part gives amplitude $e^{-x/\ell_{\rm amp}}$. Power decays twice as fast in its exponent, so its attenuation length is $\ell_{\rm amp}/2$. Finite conductivity changes the speed only at higher order in this constant-parameter weak-loss expansion, apart from any separately retained internal inductance or dielectric response.
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