Use ohm's law. The voltage drop across the resistor will be the supply
voltage (13.8) minus the discharged battery voltage (10 volts) or 3.8
volts. R = E / I so 3.8 / 8 = 0.475 ohms. So a 0.5 ohm resistor. The
power disipated is I*I*R or 8*8*0.5 = 32W.

The battery voltage will quickly climb and the charging current will
drop accordingly so the battery takes longer to charge this way but it
should be fine if you normally "float" it across the supply to keep it
charged *and* it will protect the battery and the supply from over
current when it is discharged.