= Solution
Assume the two <thermodynamic phases> have similar atomic volumes, so use $f_V\simeq0.15$. Model the stated average pitch $L=1\ {\rm \mu m}$ by one sphere per representative cubic volume $L^3$. If the sphere diameter is $d$,
$$
f_V=\frac{(\pi/6)d^3}{L^3},
\qquad
\boxed{d=L\left(\frac{6f_V}{\pi}\right)^{1/3}
\simeq0.66\ {\rm \mu m},\quad r\simeq0.33\ {\rm \mu m}.}
$$
Using the more literal lever-rule reading0.145 gives about0.65micrometres instead, within graphical uncertainty. The central estimate is therefore a diameter of roughly two-thirds of a micrometre.
A random distribution does not make a nearest-neighbor mean distance exactly equal to the cube root of inverse <number density>. The representative cubic spacing is an explicit geometric closure, not evidence that the <precipitates> actually form a cubic lattice. Unequal <molar volumes> or a different spacing statistic would change the inferred size.
\Image[/past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2015/paper-1-precipitate-estimates.png]
{title=Original geometric estimates for a spherical precipitate fraction, a dislocation free gap and a trial internal domain wall}
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