What do you mean by most efficient? All of the power of the drop will be lost unless you are willing to do a switching regulator.
The simplest way would be to us an adjustable linear regulator of the appropriate power rating. The LM317 is the jellybean part that is used for this application:
You can make a circuit with a power transistor and two resistors, to give you some fairly stable voltage drop, but it varies with temperature a little bit, and is not as stiff as using the LM317. And the two options are about the same in size.
Remember that you need to dissipate that heat somehow, and keep it away from the battery itself. Use appropriate heat sinking or some other option to get rid of the heat.
EDIT -- fixed typo "power resistor and two resistors" -> "power transistor and two resistors" (jeeze)
The application it will be used for is a bike light. 6v 15w bulb, the battery is a 10.8v 3.6 Ah that is doing nothing right now. I figure a burn time of about 2.5 hrs with this if I get it down to 6 volts. Because its on a bike I wanted to keep it small. Also thinking about a digital controller. Efficiency? I didn't want to have the 4 extra volts being turned into heat vs. getting a longer run time from the battery, but I'm not sure what is possible.