Start with a balanced equation.
#"Cu(s) + 4HNO"_3("aq")"##rarr##"Cu(NO"_3)_2("aq") + "2NO"_2("g") + "2H"_2"O(l)"#
Calculate mol #"Cu"# by dividing its given mass by its molar mass #("63.546 g/mol")#. Do this by multiplying by the inverse of its molar mass (mol/g). Then multiply by the mol ratio between #"Cu"# and #"NO"_2"# in the balanced equation, with #"NO"_2"# in the numerator. Then calculate the mass of #"NO"_2"# by multiplying by the molar mass of #"NO"_2"#, #("46.005 g/mol")#.
#8.735color(red)cancel(color(black)("g Cu"))xx(1color(red)cancel(color(black)("mol Cu")))/(63.546color(red)cancel(color(black)("g Cu")))xx(2color(red)cancel(color(black)("mol NO"_2)))/(1color(red)cancel(color(black)("mol Cu")))xx(46.005"g NO"_2)/(1color(red)cancel(color(black)("mol NO"_2)))="12.65 g NO"_2"#