Formula:1.73 X V X I/1000

180X1000/1.73X415 =I

Full load current 250A.

But we should have take only 80% 

Now a days DG manufactures assuring that the DGs can be 
loaded up to 120% of its rated capacity. But it is always 
good to design DG for a maximum load of 80% of the rated 
capacity to get the maximum efficient running. In your case 
You can load a maximum load of 125.8 kW at pf 0.85. So you 
have to give an incomer of MCCB with rating suitable for 
125.8kW.

W=root3 x VI x pf
so I = W/root3 x V x pf
I think 250A, 16kA, 4Pole MCCB with 80% plug setting may be 
suitable for you.

Accordingly select your incomer with suitable kA rating 
with respect to your total load.

BETTER AS THUMB RULE 
I = kVA X 1.4
ie 
 180 X 1.4 = 252A (AS FULL LOAD)

I = KVA/ 1.732*V
or
I = KW/ 1.732*pf*V
 
u can also use the fector as mantion Ranjith  

I = KVA * 1.4

Since the Genset Rating is given in KVA as 180KVA.

I think you should calculate the full load Amperage as 
follows:

Formula
I=KVAx1000/Root3x400V

ie, 180x1000/1.732(400)
=180,000/692.82
=259A as full load Amperage.

Generator rating is 180KVA.
Actually all electrical data is made avialable on genrator 
name plate by manufacturer.

If you consider the PF is 0.8 Lag.

Then KW=KVA*PF
i.e. KW=180*0.8
       =144 KW
P(KW)= 1.732*V*I*PF
Then,

144=1.732*415*I*0.8

I= (144/1.732*415*0.8)*1000
I= 250.42 Amps.

This is the full load current of generator with 0.8 PF & 
415V.

In case of DG sets the generator O/P power is restricted by 
power of driving engine, beyond that limit you can,t load 
the particular generator. You can put upto 125% of to 
genrator but not for too long period.

kva   =kva*1000
       -----------
       1.732*v*i*cos0

       180*1000
     =  --------
       1.732*440*i*0.85

   I  =180000
      --------=278 A
       647.768