= Solution
Interpret the given noise correlation as the three-dimensional dot-product correlation. Each Cartesian component then has $\langle\xi_i(t)\xi_j(t')\rangle=2\gamma k_BT\delta_{ij}\delta(t-t')$. For the usual Gaussian thermal-noise idealization, each coordinate is an <Ornstein-Uhlenbeck process>. With $\tau=\gamma/k$,
$$
\mathbf x(t)=\mathbf x_0e^{-t/\tau}+
\frac1\gamma\int_0^t e^{-(t-s)/\tau}\boldsymbol\xi(s)\,ds.
$$
For a particle initially at $\mathbf x_0=0$,
$$
\langle|\mathbf x(t)|^2\rangle=
\frac{6\gamma k_BT}{\gamma^2}\int_0^t e^{-2(t-s)/\tau}ds
=\frac{3k_BT}{k}(1-e^{-2t/\tau}).
$$
Hence its long-time displacement about the trap/elastic equilibrium has
$$
\boxed{x_{\rm rms}=\sqrt{3k_BT/k}}.
$$
This also follows from <equipartition theorem>: each of three quadratic coordinates contributes $k_BT/2$.
A different common experimental quantity is the stationary time-lag displacement. It obeys
$$
\langle|\mathbf x(t+\Delta)-\mathbf x(t)|^2\rangle
=\frac{6k_BT}{k}(1-e^{-\Delta/\tau}),
$$
with long-lag RMS $\sqrt{6k_BT/k}$. \b[The factor of two distinguishes fluctuations about a mean <position> from differences between two decorrelated <positions>.] The observation convention must be stated rather than silently interchanged.
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