I'm going to assume:
-- the indicator (on the dashboard) is in series with the tank unit;
-- the ideal battery is a lead-acid type, with terminal voltage of 13.6 VDC;
-- the car is wired with cryogenic superconducting wire, so there's no other resistance anywhere in the gas-gauge circuit.
Now I'll go ahead and solve the problem that I just invented.
a). when the tank is empty
-- indicator resistance = 10.9
-- tank unit resistance = 162
-- total circuit resistance = 172.9
Current in the circuit = V/R = (13.6 V) / (172.9 ) = 78.7 mA
b). when the tank is half-full
-- indicator resistance = 10.9
-- tank unit resistance = (1/2) (162 + 24.4 ) = 93.2
-- total circuit resistance = 104.1
Current in the circuit = V/R = (13.6 V) / (104.1 ) = 130.6 mA
c). when the tank is full
-- indicator resistance = 10.9
-- tank unit resistance = 24.4
-- total circuit resistance = 35.3
Current in the circuit = V/R = (13.6 V) / (35.3 ) = 385.3 mA