What volume, in drops, of concentrated HNO3 (6M) is required to react with 0.0191 g of Cu metal (20 drops per milliliter)?

The first thing that you need to do is write out the
reaction:

Cu + 4HNO3 --> Cu(NO3)2 + 2NO2 +
2H20

Next you want to convert the mass of copper to
moles:

Molar mass of copper = 63.54 g/mol
Moles =
Mass/(Molar Mass) = 0.0191g / 63.54g/mol = 0.0003
moles

Next you want to multiply by 4 to determine the
number of moles of HN03 needed: 0.0012 moles

As stated in
the question the solution is 6 moles/L or 0.006 moles / mL.  Divide that by the
20drops/mL and you have a concentration of 0.0003
moles/drop

If we divide the number of moles needed, by the
number of moles per drop we can find out how many drops of HN03 are needed: 0.0012 /
0.0003 = 4

Therefore, you need 4 drops of
HNO3