You can't disregard the amperage as Ohm's law tells you that the voltage drop that occurs across the resistor will depend upon the current. However, you could do what you want using two resistors connected in series. One 0.6 Ohm and one 2.7 Ohm connected in series. The free end of the 2.7 Ohm connected to 0V and the free end of the 0.6 Ohm connected to 3V. Then the voltage that will appear across the 2.7 ohm resistor will be 2.7 V.
If you can't find a 0.6 Ohm you can make it up by a series combination of 0.33 +0.27 Ohm.
If your deciding which resistor is to use to drop this source, amperage is an important thing. Commercially available resistors are rated in watts. Depending on the application, you would buy a higher wattage rating (btw watts or power = I^2R or square of ampere times the resistance) for higher ampere ratings.
Use a voltage regulator. It must be designed to handle the current, so you can't disregard it.
Put one diode 1N4007 cost 10 cents in series with 3.3v cell shall do.
Connection....
+3.3V terminal ---- anode from 1N4007 ----- output from 1N4007 provide about 2.7V
Which resistor should I use in order to drop my voltage from 3.3 to 2.7 volts? Disregarding amperage.
