In th figure a sphere fitted in an octahedral void is shown by shaded part

Since $∆\mathrm{ABC}$ is right angle triangle.

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{CB}}^2 \\ {\left(2\mathrm{R}\right)}^2={(\mathrm{R}+\mathrm{r})}^2+{(\mathrm{R}+\mathrm{r})}^{2}4{\mathrm{R}}^2=2{(\mathrm{R}+\mathrm{r})}^2 \\ 2{\mathrm{R}}^2={(\mathrm{R}+\mathrm{r})}^2 \\ \sqrt{2{\mathrm{R}}^2}=\sqrt{{(\mathrm{R}+\mathrm{r})}^2} \\ \sqrt{2}\mathrm{R}=(\mathrm{R}+\mathrm{r}) \\ \mathrm{r}=\sqrt{2}\mathrm{R}-\mathrm{R} \\ \mathrm{r} =0.414 \mathrm{R} \end{gathered}$$

Here $\mathrm{R}$ is the radius of atom.