Answer
$96600\,C\,mol^{-1}$
Work Step by Step
Copper is deposited according to the equation
$Cu^{2+}+2e^{-}\rightarrow Cu(s)$
$2F$ charge is required for the deposit of 1 mole Cu.
Now, $Q=It=3.00\,A\times304\,s=912\,C$
912 C charge was required to deposit 0.300 g or $\frac{0.300\,g}{63.546\,g/mol}=0.00472\,mol$ Cu
$\implies$ Charge required for the deposit of 1 mol Cu=$\frac{912\,C}{0.00472\,mol}=2F$
Or $F=\frac{912\,C}{2\times0.00472\,mol}=96600\,C\,mol^{-1}$
