about power supply on what length of copper wire we should use
The answer is 


400+ mm^2 copper made up of 50+ wires per phase conductor x 3 conductors + 185 mm^2 copper made up from 30+ wires for the neutral conductor.   See details below
If you are trying to implement this in practice and need to ask this question then your chances of dying if you try to use any answer you get is high.
If this is for an assignment then you are doing yourself a disservice by asking the question in this manner here rather than trying harder to find out for yourself by using the tools available. 


Presumably you mean "If I have a 400V, 530 kW system, what conductor diameter do I need to carry the power 350 metres.  

The answer depends on applicable regulations, load type, cable environment (buried, free air, ...) and more.  

530 kW / 400V =~ 1,320A (not 132A mentioned elsewhere.)
 This is the same as 3 x 230V phases at 530 kW / 230V / 3 phases  =~ 550 A/phase. 

This is a very serious level of power and  amount of current by any measure.  

You don't mention what sort of load you are using - if your load has a reactive component then your VA rating may be higher or much higher than your Watts and you may need larger cables again. There are many important and critical issues to answer when designing at this power level. If you do not use the services of an experienced professional then disaster is about certain.



Read this!:

This superb 70+ page document "Review of Power Cable Standard rating Methods" from a 2004 IEEE publication will provide you with a very solid grounding in what is required, and why. 



Real world product - example only:

As a guide, the very largest cable on this useful page is possibly almost large enough for your load in some cases.  The figures I use below may differ slightly from the table values. The results are close enough to provide a feel for what would be experienced. 

Losses over 350m at 500A /phase would be about 5 kW/phase or 15 kW total. Or about 3% of input power. 

Voltage drop per phase would be V=IR = 500A x 0.06 ohm = 10 Volt.

That's resistive drop. Note from the table that reactive resistance (reactance) is slightly higher than resistance for a cable this large, so that actual drop would be more like 15 Volts/phase.

The table below from here gives some sort of indication of what is required.