${\left[{\mathrm{MnCl}}_6\right]}^{-3}$ has ‘X’ unpaired electrons and ${\left[\mathrm{Mn}{\left(\mathrm{CN}\right)}_6\right]}^{-3}$ has ’Y’ unpaired electrons of the metal ion, then the ratio of X and Y is

$MnC{l_6}^{-3}$ is high spin complex with 4 unpaired electrons and ${\left[Mn{\left(CN\right)}_6\right]}^{-3}$ is a low spin complex with 2 unpaired electrons

Hence ratio $\frac{4}{2}\Rightarrow 2$