$\quad \quad \quad \quad \quad \quad x \quad \quad x$

$\mathrm{CaC}_{2} \mathrm{O}_{4} \rightarrow \mathrm{Ca}^{2+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$

$\quad \quad \quad \quad \quad \quad y \quad \quad y$

Now, $\left[\mathrm{Ca}^{2+}\right]=x+y$

and $x(x+y)=4.7 \times 10^{-9}$

$y(x+y)=1.3 \times 10^{-9}$

Dividing equation $(i)$ and $(ii)$ we get $\frac{x}{y}=3.6$

$\therefore \quad x=3.6 y$

Putting this value in equation $(ii)$, we get

$y(3.6 y+y)=1.3 \times 10^{-9}$

On solving, we get $y=1.68 \times 10^{-5}$ and $x=3.6 \times 1.68 \times 10^{-5}=6.048 \times 10^{-5}$

$\therefore \quad\left[\mathrm{Ca}^{2+}\right]=(x+y)=\left(1.68 \times 10^{-5}\right)+\left(6.048 \times 10^{-5}\right)$

$\therefore \quad\left[\mathrm{Ca}^{2+}\right]=7.728 \times 10^{-5}\, \mathrm{M}$