You need to know the VAR capability of the generator, the running load, and motor data including the starting pf of the motor. When the motor starts it will have a low pf, which means it will be drawing a lot of Vars and the generator needs to be able to supply that.
I venture to guess since you are in Liberia, at 380V 3phase, the motor is going to be an IEC design, which then Hp plays no factor, it is all KW rated. The LRA can vary greatly, I have seen them as high as 12X FLA. If you are specifying it than you can control that part to some degree.
So, lets say it is a 1800W motor (3A x 380V x 1.732 x 0.92pf) and LRA is going to be 6X, than on starting the current is 18A.
This translates to 11.85KVA. Assume a .28 starting pf and you need 11.38kVars on starting. We'll assume the running load is resistive so it will not need any Var contribution.
Your generator will need to supply running load plus starting load:
1.3KVA or kW (resistive) + starting load 3.3kW = 4.6kW
and starting Vars of 11.38kVars for total; (4.6kW +j11.38kVar) or 12.3KVA @ 68deg.
Looking at generator Var capability curve will determine final size of generator. You could be conservative and use rated standby size which is approx 10% greater than prime rating. The pf is always at 0.8 on prime rating which translates to 36.87deg.
Which means you would need a machine capable of minimum 19kVA @ 0.8pf (standby) or 17.3KVA (Prime) to meet the 11.38kVar starting requirement.
Ok, someone PLEASE check my numbers, its been a while since I did that by hand.