Answer
$\tau = 0.382~N~m$
Work Step by Step
Two-thirds of the energy (6.00 kJ/min) goes into the motor output. We can find the power of the motor output.
$P = \frac{E}{t} = \frac{6000~J}{60~s}$
$P = 100~W$
We can find the torque that the engine will develop.
$P = \tau ~\omega$
$\tau = \frac{P}{\omega}$
$\tau = \frac{100~W}{(2500~rpm)(1~min/60~s)(2\pi~rad/rev)}$
$\tau = 0.382~N~m$
