Assertion(A) : Solubility of $BaS{O}_4$  in  $0.1MN{a}_{2}S{O}_4$ is ${10}^{-9}M$  hence it $K_{SP}$  is  ${10}^{-18}$
Reason(R) : In Aqueous solution, solubility product of $BaS{O}_4$   = $S^2$  (Where S is a solubility of $BaS{O}_4$ ) 
 

$\begin{array}{l}{\text{BaSO}}_4\rightleftharpoons {\text{\hspace{0.17em}Ba}}^{2+}+{\text{SO}}_4^{-2}S={10}^{-9}\\ K_{SP}={10}^{-9}\times 0.1(0.1+{10}^{-9}\simeq 0.1)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(S+0.1)\hspace{0.17em}}\\ \therefore K_{SP}=\left[B{a}^{+2}\right]\left[S{o}_4^{-2}\right]\\ \therefore K_{SP}={10}^{-9}\times (0.1+{10}^{-9})\\ \\ K_{SP}={10}^{-10}\end{array}$

Ans  $K_{SP}\left(BaS{O}_4\right)=S^2$  (in aqueous solution)