$\mathrm{pH}$ of Urine $=6.0$

$\Rightarrow \left[{\mathrm{H}}^{+}\right]$ i.e. concentration of ${\mathrm{H}}^{+}={10}^{-6}{\mathrm{molL}}^{-1}$

$\Rightarrow$ For $1 \mathrm{L}$ of urine ${10}^{-6}$moles of ${\mathrm{H}}^{+}$ions are present

$\therefore$ If $1.3 \mathrm{L}$ of urine is eliminated, then the moles of ${\mathrm{H}}^{+}$ion eliminated will be

$1 \mathrm{L}\to {10}^{-6}$

$1.3 \mathrm{L}\to \mathrm{x}$

$x=1.3\times {10}^{-6}$

Hence the correct option is (B) $1.3\times {10}^{-6}$.