= Solution
The <Stokes–Einstein relation> gives the translational <diffusion coefficient>
$$
D=\frac{k_BT}{6\pi\eta R}=8.17\times10^{-11}\ \mathrm{m^2\,s^{-1}}
$$
at $T=298\ \mathrm K$. For isotropic three-dimensional <Brownian motion>, the <mean squared displacement> is $\langle|\Delta\mathbf r|^2\rangle=6Dt$. Thus after one hour the <root mean square> displacement is
$$
\boxed{\sqrt{6Dt}=1.33\ \mathrm{mm}.}
$$
The phrase “on average” needs a convention: the mean displacement vector is zero; the <mean displacement magnitude of three-dimensional Brownian motion> is $\langle|\Delta\mathbf r|\rangle=4\sqrt{Dt/\pi}=1.22\ \mathrm{mm}$, obtained by integrating the three-dimensional <Gaussian distribution>. These are displacements, not the total microscopic path length. The estimate assumes dilute unconfined water, a spherical hydrodynamic radius and constant <temperature>.
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