= Solution
Put <line charges> $+\lambda$ and $-\lambda$ on the two axes. For $d\gg a_1,a_2$, the surface potentials differ by
$$
V_1-V_2\simeq\frac{\lambda}{2\pi\epsilon_0}
\left(\ln\frac d{a_1}+\ln\frac d{a_2}\right).
$$
Thus the <capacitance per unit length> in vacuum is
$$
\boxed{C'=\frac{2\pi\epsilon_0}{\ln[d^2/(a_1a_2)]}
=\frac{\pi\epsilon_0}{\ln[d/\sqrt{a_1a_2}]}.}
$$
The two logarithms belong to one potential difference, not two independent <capacitances> in parallel. A homogeneous <dielectric> replaces $\epsilon_0$ by its absolute <permittivity>; finite radii and proximity corrections are neglected.
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