A mixture contains $0.100$ mol of $\mathrm{FeC}{}_2\mathrm{O}{}_4$ and $0.0500$ mol of $\mathrm{Fe}{}_2(\mathrm{C}{}_2\mathrm{O}{}_4){}_3$. Assuming only $\mathrm{Fe}{}^{2+}$ and oxalate are oxidised, the moles of $\mathrm{KMnO}{}_4$ required in acid are

Fe2+ comes only from FeC2O4: 0.100 mol, giving 0.100 mol electrons. Oxalate = 0.100 + 3 x 0.0500 = 0.250 mol, giving 0.500 mol electrons. Total = 0.600 mol e-, and KMnO4 = 0.600/5 = 0.120 mol.