if the motor is rated for 4kW at 380 volt, estimate the starting current and calculate the full load line and?

start current
p = iv
4000 = 380*i
i = 10.5A

running current is .85 of the start current or 8.9A

The full load current is determined by calculating the KVA of the motor first.
KVA=KW / PF = 4 / 0.85 = 4.706 KVA
The FLA = VA / volts =4706 / 380 =  12.38 amperes
The starting current will be approximately 600% of FLA = 6 * 12.38 = 74.3 amperes

The overload setting will depend on whether this motor is rated for a greater than FLA operation and the type overload protection you are using. If you are using an overload relay having heaters installed in the overload relay, you will have to select the heater sized on the FLA and the specific overload relay. I would use a 15 ampere circuit breaker as a minimum.

Edit:

Vshade is correct in assuming the motor to be 3-phase. My error. To correct my calculations, divide my answers by 1.732.

TexMav

Any motor with a rated voltage higher than 240 volts must be assumed to be a 3-phase motor.

Any motor rated in kW (not horsepower) should be assumed to conform to IEC standards. Local regulations may mandate which of several IEC efficiency standards applies. There may also be local regulations applicable to starting current. IEC seems to permit 800 to 1000 percent of full load current.

Input current = Output Power (Watts) / (Volts X 1.732 X efficiency X power factor)
Full load line current = 4000 / (380 X 1.732 X 0.87 X 0.85) = 8.2 amps

Starting current = 8 to 10 X 8.2 = 66 to 82 amps.

Local regulations may provide directions for selecting the overload setting based on full load line current.

PS 87% efficiency estimate is based on IEC 2 efficiency for a 4kW motor.