= Solution
Use the squared three-dimensional endpoint <displacements>, not their arithmetic <mean>. The <mean squared displacement> estimate and corresponding <diffusion coefficient> are
$$
\widehat{\langle\Delta r^2\rangle}=\frac{31^2+2^2+50^2+65^2+14^2}{5}=1577.2\,\mu\mathrm m^2,
\qquad \widehat D=\frac{1577.2}{6\times60}=4.38\,\mu\mathrm m^2\,\mathrm s^{-1}.
$$
The factor six is for three-dimensional <Brownian motion>. For an isolated no-slip sphere, the <Stokes–Einstein relation> gives the <Brownian radius estimate from endpoint displacements>:
$$
\boxed{a=\frac{k_BT}{6\pi\eta\widehat D}\simeq5.6\times10^{-8}\,\mathrm m=56\,\mathrm{nm}.}
$$
Here room <temperature> is taken as $298\,\mathrm K$; choosing $293\,\mathrm K$ gives $55\,\mathrm{nm}$, a much smaller difference than the sampling uncertainty. Under independent increments with a <normal distribution>, $\operatorname{sd}(\widehat D)/D=\sqrt{2/(3n)}\simeq0.37$ for $n=5$. The <sampling uncertainty of an endpoint diffusivity estimate> therefore makes this a rough <radius> estimate. It assumes negligible drift and hydrodynamic coupling between probes, the supplied bulk <viscosity>, and actual three-dimensional tracking. Localization noise adds to the squared <displacements> and would bias this inferred <radius> downward if uncorrected.
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