(A) Total charge passed $Q = I \times t = 10 \, A \times 193 \times 60 \, s = 115800 \, C$.
Effective charge considering $80 \%$ efficiency $= 115800 \times 0.8 = 92640 \, C$.
In terms of Faraday: $Q = \frac{92640}{96500} \approx 0.96 \, F$.
Cathode reaction: $Cu^{2+} + 2e^- \rightarrow Cu_{(s)}$.
Anode reaction: $H_2O \rightarrow \frac{1}{2} O_2 + 2H^+ + 2e^-$.
Weight of $Cu$ deposited $= \frac{0.96 \, F}{2 \, F} \times 63.5 \, g \approx 30.48 \, g$.
Moles of $O_2$ produced $= \frac{0.96 \, F}{4 \, F} = 0.24 \, \text{mol}$.
Using $PV = nRT$,$V = \frac{nRT}{P} = \frac{0.24 \times 0.08314 \times 300}{1} \approx 5.986 \, L$.