In the hydrolysis $6\mathrm{XeF}{}_4+12\mathrm{H}{}_2\mathrm{O}\to 4\mathrm{Xe}+2\mathrm{XeO}{}_3+24\mathrm{HF}+3\mathrm{O}{}_2$, the number of moles of electrons gained by Xe(IV) that is reduced to Xe(0), per 6 mol of $\mathrm{XeF}{}_4$, is

Each Xe(IV) reduced to Xe(0) gains 4 electrons, and 4 mol of Xe are formed, so $4\times 4=16$ mol of electrons are gained. Check: 2 mol of Xe(IV) go to Xe(VI), releasing $2\times 2=4$ mol e-, and 3 mol $\mathrm{O}{}_2$ release $3\times 4=12$ mol e-, a total of 16. The answer is 16.