Induction motor simple power and torque
Torque is given by:

$$
\tau = \frac{P_{shaft}}{\omega}
$$

Where \$P_{shaft}\$ is the output power of the motor (shaft power) and \$\omega\$ is the angular frequency of rotation.

The power supply in my country is 50 Hz, so a 4-pole motor has a synchronous (zero slip) speed of 1,500 RPM. Converting to angular frequency (radians per second), we get

$$ 1,500\ RPM \times \frac{2\pi\  rad.s^{-1}}{60\ RPM} = 157.1\ rad.s^{-1} $$

Slip is 2%, so the motor is running at 98% of synchronous speed :

$$
\omega = 157.1\ rad.s^{-1} \times 0.98 = 153.9\ rad.s^{-1}
$$

The output power of your motor is 100 HP. In metric units that is

$$
P_{shaft} = 100\ HP \times \frac{746\ W}{1\ HP} = 74,600\ W
$$

Giving

$$
\tau = \frac{74,600\ W}{153.9\ rad.s^{-1}} = 484.6\ N.m
$$

Note I have written down the units of every figure I have used. This makes it much easier to spot mistakes.