= Solution
At maximum projected separation, an Earth analogue at $10\,\mathrm{pc}$ is separated from its star by $\vartheta=a/D_{\rm sys}=0.1\,\mathrm{arcsec}=4.85\times10^{-7}\,\mathrm{rad}$. The <Rayleigh criterion> for a circular pupil gives
$$
\boxed{D_{\rm tel}\gtrsim\frac{1.22\lambda}{\vartheta}\simeq1.26\,\mathrm m\quad(0.5\,\mu\mathrm m),\qquad50.3\,\mathrm m\quad(20\,\mu\mathrm m).}
$$
The relevant quantity is the star–planet separation, not the tiny angular diameter of the planet itself. Simple angular <optical resolution> is insufficient: the planet lies in the star's <diffraction> pattern at reflected <contrast> of order $10^{-10}$, or thermal <contrast> of order $10^{-6}$. Residual speckles, wavefront errors, scattered light, detector/background noise and, in the infrared, strong telescope and sky thermal emission obstruct detection. A <coronagraph> or other stellar suppression method generally has an inner working angle of several $\lambda/D$, making practical aperture requirements larger than the bare <Rayleigh criterion>. The question's instrumentation claim is interpreted in its 2019 context; the physical requirements remain the basis of the estimate. Maximum reflected brightness near full phase also coincides with small projected separation, so maximum brightness and maximum resolvable separation cannot generally be obtained simultaneously.
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