${\mathrm{HNO}}_3$   oxidizes ${\mathrm{NH}}_4^{+}$ ions to nitrogen and itself gets reduced to ${\mathrm{NO}}_2.$

The moles of ${\mathrm{HNO}}_3$ required by 1 mole of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{is}$

$\begin{array}{l}\mathrm{H}\stackrel{+5}{\mathrm{N}}{\mathrm{O}}_3+\stackrel{-3}{\mathrm{N}}{\mathrm{H}}_4^{+}\to {\stackrel{0}{\mathrm{N}}}_2+\stackrel{+4}{\mathrm{N}}{\mathrm{O}}_2\\ \mathrm{V}.\mathrm{F}.\mathrm{of}{\mathrm{HNO}}_3=\left(5-4\right)=1\\ \mathrm{V}.\mathrm{F}.\mathrm{of}{\mathrm{NH}}_4^{+}=\left[0-\left(-3\right)\right]=3\\ \mathrm{so}\mathrm{molar}\mathrm{ratio}\mathrm{of}{\mathrm{HNO}}_3\mathrm{and}{\mathrm{NH}}_4^{+}\mathrm{is}3:1.\\ 1\mathrm{mole}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\mathrm{is}\mathrm{found}\mathrm{to}\mathrm{contain}2\mathrm{mole}\mathrm{of}{\mathrm{NH}}_4^{+}\\ \mathrm{So},\mathrm{required}\mathrm{moles}\mathrm{of}{\mathrm{HNO}}_3\mathrm{is}3\times 2=6\mathrm{moles}.\end{array}$