The solubility of ${\mathrm{BaSO}}_{4 }$$(\mathrm{molar} \mathrm{mass}= 233 \mathrm{g} {\mathrm{mol}}^{-1})$ in $100 \mathrm{mL}$ solution is $9.32 \mathrm{x} {10}^{-4} \mathrm{g}.$ Its solubility in $0.1 \mathrm{M} {\mathrm{Na}}_2 {\mathrm{SO}}_4$ solution will be

Solubility (in  mol/L) is $s=\left(\frac{9.32\times {10}^{-4}\mathrm{g}}{233\mathrm{g}{\mathrm{mol}}^{-1}}\right)\left(\frac{1000{\mathrm{mLL}}^{-1}}{100\mathrm{mL}}\right)=4\times {10}^{-5}{\mathrm{molL}}^{-1}$

For the equilibrium ${\mathrm{BaSO}}_4\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Ba}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right),K_{\mathrm{s}}\left({\mathrm{BaSO}}_4\right)=\left[{\mathrm{Ba}}^{2+}\right]\left[{\mathrm{SO}}_4^{2-}\right]=s^2=\left(4\times {10}^{-5}{\mathrm{molL}}^{-1}{)}^2\right.$Solubility of ${\mathrm{BaSO}}_4$in $0.1\mathrm{M}{\mathrm{Na}}_2{\mathrm{SO}}_4$ is $\left[{\mathrm{Ba}}^{2+}\right]=\frac{K_{\mathrm{s}}\left({\mathrm{BaSO}}_4\right)}{\left[{\mathrm{SO}}_4^{2-}\right]}=\frac{{\left(4.0\times {10}^{-5}{\mathrm{molL}}^{-1}\right)}^2}{\left(0.1{\mathrm{molL}}^{-1}\right)}=1.6\times {10}^{-8}\mathrm{M}$