= Solution
Use only the high-pressure <CsCl structure> branch to extrapolate its <volume> per formula unit $V=a^3$. A straight-line least-squares fit to those three measurements gives approximately
$$
V(P)=(52.285-1.300P)\,\mathrm{\mathring A^3},
$$
with $P$ in GPa. Thus $V(10)\simeq39.29\,\mathrm{\mathring A^3}$, $a(10)\simeq3.400\,\mathrm{\mathring A}$ and $d_{111}=a/\sqrt3\simeq1.963\,\mathrm{\mathring A}$. Taking Cu K$\alpha$ wavelength $\lambda\simeq1.541\,\mathrm{\mathring A}$ in <Bragg's law> gives
$$
\boxed{\theta\simeq23.1^\circ,\qquad 2\theta\simeq46.2^\circ}.
$$
The estimate is model-dependent: a straight-line extrapolation of $a(P)$ instead gives $2\theta\simeq46.0^\circ$. The data stop at $3.81\,\mathrm{GPa}$, so the absence of further phase transitions does not by itself determine an exact equation of state to $10\,\mathrm{GPa}$. The $111$ peak can also be weak because its basis factor is $f_K-f_{Cl}$.
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