I would use a pot of say 50K ohms. The run the voltage through the pot and the wiper will be the voltage out. It is the same as a voltage divider circuit but uses the wiper to make two resistors, one on each side of the wiper.
Something like this:
-0 | 0-
the circles are the terminals to the pot and the | is the wiper. This is a general solution that can be used whenever you need to get less voltage then you have.
Use your VOM to see what the voltage is on the wiper and you are done. If you want to check it you can then test the resistance between each terminal and the wiper. And, with nothing up my sleeve, I predict that you will find that the ratio is the same as what the voltage divider formula predicted.