It's 80.1% for the copper purity sample.
I'm writing up solutions here - this is a tricky question, but my reasoning is here:
"This question is actually fairly tricky. What I gather is that the mass of copper in the impure electrode is more accurately reflected by the amount of copper that has been deposited on the cathode. The reason for this is because the copper deposited on the cathode is a result of the amount of electrons pulled away at the anode by the current. Metals that are more reactive than copper would all dissolve in the H2SO4, and that would account for some of the mass loss in the impure sample. Metals that are less reactive than copper will be, as stated in the following part, not oxidised and will form a sludge."
If you check the electrochemical series, copper is the most reactive element that does not react with H+.
@a-rod: If there is copper left in the impure sample after electrolysis, then determining the percentage purity of copper in the original impure sample would be inaccurate.