From 5v to 3.3v at from 8mA to 160mA
Let's go through your list of options.


Voltage divider wouldn't work.  Usually, the textbook voltage divider assumes that the output voltage is measures with an open circuit.  In you case, there will be a load with a varying current.  Output voltage of the divider will depend on the load and will not be regulated.
Series diodes.  You have already found the weakness of this approach: regulation is not very good.
Voltage regulator is the right method for doing this.  If you use a linear regulator, then the power dissipation will be P = (5V - 3.3V) * 0.16A = 0.272 W.  This is manageable.  TO-92 package, for example, can dissipate 0.5 W.
A switch-mode step down regulator (buck) will dissipate even less.