${\mathrm{HNO}}_3$ 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 ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$ is

$$\begin{gathered} \mathrm{H}\stackrel{+5}{\mathrm{N}}{\mathrm{O}}_3 + \stackrel{-3}{\mathrm{N}}{\mathrm{H}}_4^{+} \stackrel{ }{\to } {\stackrel{0}{\mathrm{N}}}_2 + \stackrel{+4}{\mathrm{N}}{\mathrm{O}}_2 \\ \mathrm{Meq}. \mathrm{of} {\mathrm{HNO}}_3 = \mathrm{Meq}. \mathrm{of} {\mathrm{NH}}_4^{+} \\ \mathrm{moles}\times \mathrm{n}-\mathrm{factor} = \mathrm{moles}\times \mathrm{n}-\mathrm{factor} \\ \mathrm{moles}\times 1 = 1\times 3 \\ \mathrm{moles} \mathrm{of} {\mathrm{HNO}}_3 \mathrm{required} \mathrm{by} 1 \mathrm{mole} \mathrm{of} {\mathrm{NH}}_4^{+} = 3 \\ \mathrm{moles} \mathrm{of} {\mathrm{HNO}}_3 \mathrm{required} \mathrm{by} 1\mathrm{mole} \mathrm{of} {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4 = 6 \end{gathered}$$