How to find the concentration of barium ions in a mixture of barium sulfate and barium thiosulfate?
This is just one possible approach:

First of all, the constants doesn't seem different enough to argue that $[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]$. So for now consider:

$$[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]+[\ce{SO_4}^{2-}] \hspace{2em} (1)$$

and dividing both $K$s we can find: $$\frac{K_{\ce{BaS2O3}}}{K_{\ce{BaSO4}}}=\frac{[\ce{S_2O_3}^{2-}]}{[\ce{SO_4}^{2-}]}=4.44 \hspace{2em} (2)$$ 

Now we can solve for sulphate: $$K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}=[\ce{SO_4}^{2-}]^2(1+\frac{[\ce{S_2O_3}^{2-}]}{[\ce{SO_4}^{2-}]}) \hspace{2em} \mathrm{using}\hspace{1em}(1)$$

Using equation $(2)$ we get $[\ce{SO_4}^{2-}]=4.03\times 10^{-6}$

And for thiosulphate: $$K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}=[\ce{S2O_3}^{2-}]^{2}(1+\frac{[\ce{SO4}^{2-}]}{[\ce{S2O_3}^{2-}]}) \hspace{2em} \mathrm{using}\hspace{1em}(1)$$

Using inverse of equation $(2)$ leads to $[\ce{S_2O_3}^{2-}]=1.8\times 10^{-5}$

The sum yields 
$[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]+[\ce{SO_4}^{2-}]=0.00002215015=\sqrt(4.906295\times 10^{-10})$