= Solution
In the continuum cylindrical-pore estimate, the <electrical resistance> is $R_p=\rho\ell/(\pi a^2)$, with radius $a=0.075\,\mathrm{nm}$, length $\ell=0.5\,\mathrm{nm}$ and <electrical resistivity> $\rho=80\,\Omega\,\mathrm{cm}=0.80\,\Omega\,\mathrm m$. Therefore
$$
R_p=2.26\times10^{10}\,\Omega,\qquad
\boxed{g_p=R_p^{-1}=44.2\,\mathrm{pS}}.
$$
Unrestricted approach from the reservoirs removes access barriers, but still permits a continuum spreading or <access resistance of a circular pore>. Including $\rho/(4a)$ on each side gives
$$
R_{\rm access}=\frac{\rho}{2a}=5.33\times10^9\,\Omega,\qquad
\boxed{g_{\rm total}=\left[\frac{\rho\ell}{\pi a^2}+\frac{\rho}{2a}\right]^{-1}\simeq35.8\,\mathrm{pS}}.
$$
Thus the usual estimate neglecting the access contribution is about $44\,\mathrm{pS}$, or about $36\,\mathrm{pS}$ with spreading resistance. A $1.5$-ångström diameter is at the molecular scale, smaller than hydrated <ions>; bulk <electrical resistivity> cannot quantitatively describe dehydration, selectivity or discrete-ion transport. These values are the requested continuum estimates, not evidence that such a pore freely passes hydrated <potassium> <ions>.
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