DIY water tank gauge using analogue ammeter
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. An imbalanced bridge solution. R5 is the assumed resistance of your milliammeter.

I offer this as a rough and ready solution to your question. It is based on the principle of the Wheatstone Bridge which, when the ratios of R1:R2 = R3:R4 are matched will have zero current through R5 and AM1. R3 is set at 240 &Omega; so that balance occurs when the tank is empty and the meter will read zero.

The values for R2 and R4 were established by messing around with the CircuitLab simulator (which you can do by editing my question - just don't save it) rather than going through all the maths.

To keep the values of R2 and R4 reasonably low and avoid you having to do precise resistor matching we need to drop the bridge voltage down to about 2 V or so. R6 and D1 do this. The forward voltage, Vf, of an LED is reasonably constant over a range of currents so this voltage will remain reasonably constant with variation in battery voltage. The 680 &Omega; resistor will give about 15 mA. A red, yellow or green LED will give around 2 V at this current. Blue and white would give a higher voltage.



Figure 2. LED I-V curves for various colours of LED. Green will give a 2 V drop at around 15 mA. Source: LEDnique.



Update:

For a 1 mA, 200 &Omega; meter R2 and R4 should be reduced to about 680 &Omega;.