Passive voltage drop from 6V DC to 5V DC for 3W circuit
For any application, this very basic (and not much reliable) "dc-dc step down", could be done using a resistor between your 6V output and Vin. Considering:

Vout = 6V
Pout = 3W
Iout = Pout/Vout = 0.5A


for a total voltage drop of Vr = 1V, the resistor value should be:

R &lt;= Vr/Iout = 2 Ohm


Higher value of R won't let you use the total 3W of your power source. If value of R is too low, your circuit will have high Ibias when in standby mode (arduino not connected). Consider the case where R-->0, and see that your 5V zenner will drain more current, and also make your 6V power source more unstable.

At the same time, you should take into consideration the resistor's dissipation power:

Pr = ((Vr)^2)/R = ((1V)^2)/(2ohm) =~ 500mW


where Vr = Vout-Vin.

Finally, calculate the zener's dissipation power.

edit: for a 500mW resistor, it would be safer to use a 1W resistor. This way it won't overheat, reducing it's variation and hence the output stability.