In acidic medium $\mathrm{Cu}{}_2\mathrm{S}$ is oxidised by nitrate to $\mathrm{Cu}{}^{2+}$ and $\mathrm{SO}{}_4{}^{2-}$ while nitrate is reduced to NO. The number of moles of $\mathrm{NO}{}_3{}^{-}$ consumed per mole of $\mathrm{Cu}{}_2\mathrm{S}$ is

Per $\mathrm{Cu}{}_2\mathrm{S}$: 2 Cu go +1 to +2 (2 electrons) and S goes -2 to +6 (8 electrons), total 10. Each $\mathrm{NO}{}_3{}^{-}$ gains 3, so 10/3 mol of nitrate is needed. The full equation is $3\mathrm{Cu}{}_2\mathrm{S}+10\mathrm{NO}{}_3{}^{-}+16\mathrm{H}{}^{+}\to 6\mathrm{Cu}{}^{2+}+3\mathrm{SO}{}_4{}^{2-}+10\mathrm{NO}+8\mathrm{H}{}_2\mathrm{O}$, giving 10/3 mol per mol of $\mathrm{Cu}{}_2\mathrm{S}$.