I'm not entirely sure how to do this either (please let me know if you have the markscheme!!) but I tried working it out, and here's what I did.
So the question states to find out the value of y in the formula KH3(C2O4)y.zH2O (we can ignore the hydrated part because that's in the next question)
They've also given us the equation of the titration with Potassium Permanganate.
2MnO4- + 5C2O4(2-) + 16H+ -> 2Mn(2+) + 10CO2 + 8H2O
From the question, we know that the concentration of MnO4(-) is 0.0203mol/dm^3
The titre of MnO4(-) given to us is 23.15cm^3 --> 0.02315dm^3
Thus we can work out the moles of MnO4(-), which is 0.0203 * 0.02315 = 4.69945*10^(-4) moles
We're also given the concentration of the salt solution (containing C2O4(2-)) which is 0.0235mol/dm^3 and the volume of the solution used to titrate with the manganate, which is 25cm^3 --> 0.025dm^3
If we calculate the moles from the concentration and volume we get 0.0235 * 0.025 = 5.875*10^(-4) moles
If you notice, the MnO4 and C2O4 are in a 2:5 molar ratio but their mole values are not.
If you divide the moles of MnO4 by 2 (as it is 2MnO4(-) in the equation) you get 2.35*10^(-4) moles
then times by 5 to get moles of C2O4 (by molar ratio of 2:5) to get 1.175*10(-3) moles
lastly, if you divide the moles of C2O4 you calculated now by the moles of C2O4 we calculated earlier, you get (1.175*10(-3) / 5.875*10(-4) = 2.
therefore the value of y must be 2
you can cross check this because the charges on the original salt match up.
KH3(C2O4)2
KH3 formed 4+ and C2O4 is a 2- ion compound so if y=2, then (C2O4)2 is -4 and they balance out.
Hope that helped!! (and hopefully it is right otherwise I'm just purely embarrassed for writing all of this)