I have no idea what the actual question is, but I'm assuming that you had a 10 mL beer sample, which you placed in a 250.0 mL volumetric flask, which you dilated to 250.0 mL, from which you then withdrew a 20.00 mL aliquot, with which you then performed some sort of reaction. The question, I'm assuming, is calculate the amount of ethanol in the original 10 mL beer sample.
In part b) of the question, you found that the amount of ethanol in the 20.00 mL aliquot is 0.00068925 mol. Now, you withdrew this 20.00 mL aliquot from the 250.0 mL volumetric flask. Logically, then, the amount of ethanol in the 250.0 mL volumetric flask is 250.0/20.00 multiplied by the amount of ethanol in the 20.00 mL aliquot, which works out to be 0.008615 mol. Now we have the amount of ethanol in the 250.0 mL volumetric flask. Now, all the ethanol in the volumetric flask was sourced from the beer sample; you placed the beer sample into the volumetric flask and then added distilled/deionised water until the volumetric flask was filled up to the calibration mark. Logically, then, the amount of ethanol in the beer sample would equal to the amount of ethanol in the 250.0 mL volumetric flask. Hence, the amount of ethanol in the beer sample is 0.008615 mol, which, to three sig figs, is 0.00862 mol.
Let me know if any of the assumptions I made above are incorrect.