= Solution
Treat the given orbital separation as the projected separation on the sky. The small <angular separation> is
$$
\theta\simeq\frac{1.5\times10^{11}}{2\times10^{17}}=7.5\times10^{-7}\,\mathrm{rad}.
$$
For an ideal circular <telescope> aperture of diameter $D$, the <Rayleigh criterion> requires $1.22\lambda/D\leq\theta$. Therefore
$$
\boxed{D\geq\frac{1.22(500\times10^{-9})}{7.5\times10^{-7}}\,\mathrm m\simeq0.81\,\mathrm m.}
$$
This is the <diffraction> limit, assuming comparable image brightness and adequate sampling. Atmospheric image blurring can require correction or space-based observing. A smaller actual projected separation of the <binary star> would require a larger aperture.
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