= Solution
The fluorine sites form a simple cubic sublattice of spacing $\ell=a/2=3.10\ \mathring{\mathrm A}$. Adopt uncorrelated nearest-neighbour hopping, identify the field-correlation time with the total jump waiting time, and use homonuclear dipolar relaxation: $f_D=c=1$ and $x_*=0.616$. Then
$$
\omega_0=(2.518\times10^8)(0.25)=6.295\times10^7\ \mathrm{s^{-1}},\quad
\tau_c^*=\frac{0.616}{\omega_0}=9.79\times10^{-9}\ \mathrm s,
$$
$$
\boxed{D(T^*)=\frac{(3.10\times10^{-10})^2}{6\tau_c^*}\simeq1.64\times10^{-12}\ \mathrm{m^2\,s^{-1}}},\qquad
\boxed{D_0=D(T^*)\exp\!\left[\frac{1.52}{(8.61733\times10^{-5})(926)}\right]\simeq3.1\times10^{-4}\ \mathrm{m^2\,s^{-1}}}.
$$
If the simplified single-frequency condition $\omega_0\tau_c=1$ is used instead, the corresponding estimates are $1.01\times10^{-12}\ \mathrm{m^2\,s^{-1}}$ and $1.9\times10^{-4}\ \mathrm{m^2\,s^{-1}}$. These are explicit model-dependent estimates, not a uniquely measured tracer diffusivity: a different correlation-time-to-jump-time conversion or defect correlation factor multiplies them by $f_Dc$. The <lattice constant> supplies a jump distance, not proof that all possible neighbouring sites are vacant.
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