1 MOLE OF FERRIC OXALATE  IS OXIDISED BY X MOLE OF MNO4 MINUS  AND ALSO 1 MOLE OF FERROUS OXALATE IS OXIDISED BY Y MOLE OF MNO4 MINUS IONS IN ACIDIC MEDIUM .THE RATIO OF X/Y IS

Dear student

For this question, you need to use the concept of change in the oxidation number for the species getting oxidised or reduced. 
Also you need to use the concept of equivalents according to which the number of equivalents of reducing agent is equal to the number of equivalents of oxidising agent.
In this question one needs to take into account the change in the oxidation state for Fe and C in the different compounds. In the case of Fe₂(C₂O₄)₃, carbon is present in +3 which is getting changed to +4 oxidation state in CO₂. Also in the case of Mn, it is getting reduced from +7 oxidation state to +2 oxidation state.
Fe₂(C₂O₄)₃ + MnO₄^- -----Mn²⁺ + CO₂
The n factor of Fe₂(C₂O₄)₃  can thus be taken as equal to the n factor of MnO₄^-
Thus
Reactant (n$\times$nf) = Product (n$\times$nf)
1$\times$6= x$\times$5
Thus x = (6/5)
Now in the case of ferrous oxalate Fe(C₂O₄)
Now the new reaction can be written as:
MnO₄^-  + FeC₂O₄ ---------Mn²⁺ +Fe³⁺ +CO₂
Here also Mn is getting reduced from +7 oxidation state to +2 oxidation state.
Also in the compound, ferrous oxalate, Fe is in +2 oxidation state and C is in +3 oxidation state, which are getting changed to +3 and +4 resp.
So the total n factor for the compound is 1+2=3
Again keeping the equivalents of the reactants and the products to be same on both the sides,
Y$\times$5=1$\times$3
Thus y = (3/5)
So the ratio, (x/y) will be (6$\times$5)/(5$\times$3)
= 2:1
 



Regards