Is this for calibrating a thermocouple, for example?
I have done something similar where I needed a calibrated signal of 15.000mV. I started with a fresh Alkaline cell (I think I used a D-cell). I measured the voltage at 1.540V. A voltage divider will work for this:
Say that the upper resistor is R1 and the lower one is R2. Vout = Vin*R2/(R1+R2)
or, in my case, 0.015 = 1.54*R2/(R1+R2).
or R2/(R1+R2) = 0.015/1.54 = 0.00974
Now, for my example, I knew that the input impedance of where the output was going to be connected was very high (LCD Voltmeter), so I could pretty much pick any value of resistors without regard to loading the voltage divider. If you know what impedance the output voltage is fed to, you may have to consider that.
Back to my example. To not appreciably load the battery, I choose R1 to be 100K.
That allows me to solve for R2. Namely:
R2/(100000+R2) = 0.00974
R2 = 0.00974(100000 + R2)
R2 = 974 + 0.00974R2
R2 = 974/(1-0.00974) = 983Ω
Now 100K is a standard value, but 983 is not. However, if you break 983 into two 1% resistors, you can buy a 909Ω and a 75Ω. So three 1% 1/4W metal film resistors will do it...
Do the math for your specific example.
Note that you could just simplify the math thusly:
The current through the voltage divider is 99.99% determined (ignoring R2) by R1.
To get 1mV from 3V, and say you are willing to have 1mA flowing in the divider, then:
R1 = E/I = 3/0.001 = 3000.
Now to get 1mV with 1mA, R2 = E/I = 0.001/0.001 = 1Ω
So R1=3KΩ and R2=1Ω will work; so would 30KΩ and 10Ω, or 300KΩ and 100Ω, but that brings back the input impedance question?