Correct way of calculating BLDC torque for a given RPM and Current
The 24 N.m figure implies a mechanical power output of 1930W and efficiency of only 32%,  which is much too low if the motor is anywhere near 81% efficient at 140A. The error may be due to an incorrect assumption about phase current. 

Motor current is only the same as controller input current when the motor is getting full input voltage. If the controller limits current by reducing motor voltage (which seems to be the case here) then the motor current will be higher than the supply current. Why? Power = voltage x current. Since (assuming negligible controller loss) input and output power are equal, as output voltage is lowered so output current must increase by the same ratio.  

Motor current can be derived from Kt, rpm, and efficiency. 6034W in x 81% = 4888W out, or 61 N.m x 769rpm. Kt is the inverse of Kv, ie. 8.3/49 = 0.17 N.m/A, so motor current should be 61/0.17 = 359A, 2.56 x higher than the power supply current. 

To check this figure we can calculate the expected motor voltage and derive controller input current from the voltage reduction ratio. Voltage drop inside the motor should be (neglecting inductance) 0.011&ohm; x 359A = 4V. 769rpm/49rpm/V = 15.7V back-emf, + 4V = 19.7V applied motor voltage. 43.1V/19.7V = 2.19, so controller input current should be 359A / 2.19 = 164A. That's only 17% higher than the measured current, which means we are 'in the ballpark'.