Wiring up a AAA power toy to Pinball lighting supply
The following circuit should do what you need:



It is based on the LM317 linear regulator, which is not as efficient as a switching regulator but easier to use in a situation like yours.

The output voltage setting is determined by the two resistors:

$$V_{OUT} = 1.25V \times (1 + \frac{R2}{R1}) = 1.25V \times (1 + \frac{390}{270}) = 3.05\,V$$

Using a full wave bridge rectifier made up of 1N4007 diodes (any of the 1N400x series will do), the 6V DC coming in is rectified to a level of 6V * 1.414 = 8.5V.  Less two diode drops of at least 0.75V each, the input to the regulator will be about 7.0V or a little less.  The power dissipated in the LM317 will thus be:

$$(7.0V - 3.0V) \times 0.090\,A = 0.36\,W$$

or 360 mW.  You should not need a heat sink since this is less than a watt.

The LM317 and 1N400x series diodes are available at Digi-Key, and also RadioShack stores.