0 mL of 0.050 M Ba ( NO_3)_2 is mixed with 25.0 mL of 0.020 M NaF. K_sp of BaF_2 is 0.5 × 10^-6 at 298 K. The ratio of [ Ba^2+][ F^-]^2 and K_sp is…

$\displaystyle 0$ mL of $\displaystyle 0.050 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_3\right)_2$ is mixed with $\displaystyle 25.0$ mL of $\displaystyle 0.020 \mathrm{M} \mathrm{NaF} . \mathrm{K}_{\mathrm{sp}}$ of $\displaystyle \mathrm{BaF}_2$ is $\displaystyle 0.5 \times 10^{-6}$ at $\displaystyle 298$ K . The ratio of $\displaystyle \left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^2$ and $\displaystyle \mathrm{K}_{\mathrm{sp}}$ is $\displaystyle \_\_\_\_$. (Nearest integer)