 Solution

Assume the two thermodynamic phases have similar atomic volumes, so use $f_V\simeq 0.15$. Model the stated average pitch $L=1\mu \mathrm{m}$ by one sphere per representative cubic volume $L^3$. If the sphere diameter is $d$,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.