According to Kohlrausch's law

$$\begin{gathered} {{\mathrm{\lambda }}^{°}}_{\mathrm{m}}\left({\mathrm{BaSO}}_4\right) = {{\mathrm{\lambda }}^{°}}_{\mathrm{m}}\left({\mathrm{Ba}}^{2+}\right)+{{\mathrm{\lambda }}^{°}}_{\mathrm{m}}\left({{\mathrm{SO}}_4}^{2-}\right) \\ {{\mathrm{\lambda }}^{°}}_{\mathrm{m}}\left({\mathrm{BaSO}}_4\right) = 110+136.6 \\ {{\mathrm{\lambda }}^{°}}_{\mathrm{m}}\left({\mathrm{BaSO}}_4\right) = 246.6 {\mathrm{ohm}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

Now, the solubility product of ${\mathrm{BaSO}}_4$ can be determined by:

$$\begin{gathered} {\mathrm{K}}_{{\mathrm{BaSO}}_4} = {\mathrm{K}}_{{\mathrm{BaSO}}_4}\left(\mathrm{solution}\right)-{\mathrm{K}}_{\mathrm{water}} \\ {\mathrm{K}}_{{\mathrm{BaSO}}_4} = 3.648\times {10}^{-6}-1.25\times {10}^{-6} \\ {\mathrm{K}}_{{\mathrm{BaSO}}_4} = 2.398\times {10}^{-6} \mathrm{S} {\mathrm{cm}}^{-1} \end{gathered}$$

Finally, the molar conductivity will be:

$$\begin{gathered} {{\mathrm{\lambda }}^{°}}_{\mathrm{m}} = \frac{\mathrm{K}\times 100}{\mathrm{Solubility}} \\ {{\mathrm{\lambda }}^{°}}_{\mathrm{m}} = \frac{2.398\times {10}^{-6}\times 100}{246.6} \\ {{\mathrm{\lambda }}^{°}}_{\mathrm{m}} = 9.72\times {10}^{-6}\times 233 \\ {{\mathrm{\lambda }}^{°}}_{\mathrm{m}} = 2.26\times {10}^{-3} \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$