 Solution

Take $a,b,c>0$ as the semiaxis lengths. For a centered axis-aligned box with positive half-edge lengths $x,y,z$, the volume is $V=8xyz$. Maximize $xyz$ under $g=x^2/a^2+y^2/b^2+z^2/c^2=1$. The Lagrange multiplier equations areMultiplying by $x,y,z$, respectively, shows that $x^2/a^2=y^2/b^2=z^2/c^2$. The constraint makes each value $1/3$. All boundary boxes with a zero edge have zero volume, while the positive stationary box exists. Compactness therefore gives the maximum box volume in an ellipsoidAllowing a displaced or differently oriented rectangular box cannot improve this result. Let $A=\mathrm{diag}(a,b,c)$, let $c_0$ be its center, and let $u_1,u_2,u_3$ be its perpendicular half-edge vectors. Averaging $|A^{-1}(c_0+\sum {\epsilon }_i{u}_i){|}^2\le 1$ over the eight signs gives $|A^{-1}{c}_0{|}^2+\sum |A^{-1}{u}_i{|}^2\le 1$. The determinant bound $|\det (A^{-1}{u}_1,A^{-1}{u}_2,A^{-1}{u}_3)|\le \prod |A^{-1}{u}_i|$, followed by the arithmetic-geometric mean inequality, makes its volume at most $8abc/(3\sqrt{3})$. The box found above attains the bound.