What amount of current, in amperes, would be required to deposit 19.7 g of Au metal in 1.00 hour from a solution of Au(NO3)3?

Answer 1

You can deposit 196.97/3 = approx 66 g Au with 96,485 coulombs of electricity. So how much will be required to deposit 19.7g
96,485 x (19.7/66) = approx 29,000 coulombs.
coulombs = amperes x seconds.
Substitute coulombs and seconds and solve for amperes.
Note those numbers I used are approximate. You need to redo all of them.