A sparingly soluble salt of solubility ‘S’ has the formula $T{l}_{6}Al{\left(P{O}_4\right)}_3$ and has a solubility product of $k{S}^{2y}$ where k is a constant. The value of ‘y’ is [Thallium (Tl) is univalent ion and Aluminium (Al) is trivalent ion

  $Ba{F}_2\left(s\right)\to B{a}^{2+}\left(aq\right)+2F\left(aq\right)$
Let ‘S’ be the solubility of $Ba{F}_2$, but $B{a}^{2+}$ reacts with $C_2{O}_4^{2-}$ present in this solution &  almost completely convert into $Ba{C}_2{O}_4$
 $B{a}^{2+}+C_2{O}_4^2\to Ba{C}_2{O}_4\left(s\right)$
Let y mole per it of $B{a}^{2+}$ is left after reaching equilibrium
So,    $B{a}^{2+}+C_2{O}_4^2\stackrel{ }{\to }Ba{C}_2{O}_6\left(s\right)keq={10}^{-1}$
  $y0.1-s$
So, we get the following two questions,
 $y(0.1-s)={10}^{-2}\&y{\left(2s\right)}^2={10}^{-4}=\left(K_{hp}\right)Ba{F}_2$
Solving there two equations,
We get,  $S=0.096M$
So,  $\left(C_2{O}_4^{2-}\right)=0.1-0.096=4\times {10}^{-3}M$
 $\left[F\right]=2s=2\times 0.096-0.192M$
 $\left[B^{2+}\right]=y=2.7\times {10}^{-5}M$