Let the molar solubility of $\text{BaSO}_4$ in $0.1\text{ M Ba(NO}_3\text{)}_2$ solution be $S'$. The dissociation of $\text{BaSO}_4$ is: $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$

In the solution, $\text{Ba(NO}_3\text{)}_2$ dissociates completely to give $0.1\text{ M Ba}^{2+}$ ions. Due to the common ion effect, the total concentration of $\text{Ba}^{2+}$ ions will be: $[\text{Ba}^{2+}] = S' + 0.1 \approx 0.1\text{ M}$ (since $S'$ is very small compared to $0.1\text{ M}$)

The concentration of $\text{SO}_4^{2-}$ ions is: $[\text{SO}_4^{2-}] = S'$

The solubility product expression ($K_{sp}$) is: $K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$ $1.5 \times 10^{-9} = (0.1)(S')$ $S' = \frac{1.5 \times 10^{-9}}{0.1} = 1.5 \times 10^{-8}\text{ M}$

Thus, the molar solubility of $\text{BaSO}_4$ in the given solution is $1.5 \times 10^{-8}\text{ M}$.