Concentration of $H_{2}S{O}_4$ and $K_{2}S{O}_4$ in a solution is 0.1 M and $2\times {10}^{-3}M$ respectively. Molar solubility of $BaS{O}_4$ in the same solution is $X\times {10}^{-Y}$ M (expressed in scientific notation). The value of Y is ______.
[Given: $K_{sp}\left(BaS{O}_4\right)=3.0\times {10}^{-10};K_{a1}\left(H_{2}S{O}_4\right)={\text{very large; Ka}}_2\left(H_{2}S{O}_4\right)=1.2\times {10}^{-2}$  ]
 

$\begin{array}{l}HS{O}_4^{-}\stackrel{ }{\rightleftharpoons }{H}^{+}+S{O}_4^{2-}\\ 0.1\text{ 0.1 }2\times {10}^{-3}\\ 0.1-x0.1+x2\times {10}^{-3}+x\\ 1.2\times {10}^{-2}=\frac{(0.1+x)(2\times {10}^{-3}+x)}{(0.1-x)}\Rightarrow x\approx 1.35\times {10}^{-2}\\ BaS{O}_4\left(s\right)\stackrel{ }{\rightleftharpoons }B{a}^{2+}+S{O}_4^{2-}\\ s\text{ 1.55}\times {10}^{-2}+s\\ s\text{ 1.55}\times {10}^{-2}[\because s\text{\hspace{0.17em}is neglegible]}\\ 1.55\times {10}^{-2}s=3\times {10}^{-10}\Rightarrow s=1.9\times {10}^{-8}\end{array}$