With this logic:
GTX280 = 1400 million transistors 576mm2 at 65nm
GTX285 = 1400 million transistors 470mm2 at 55nm
4870 = 956 million transistors 260mm2 at 55nm
http://www.anandtech.com/video/showdoc.aspx?i=3405
5870 = 2,15 billion transistors 334mm2 at 40nm
http://www.anandtech.com/video/showdoc.aspx?i=3643&p=1
5870 has
2,249X (2,15/0,956) more transistors than 4870.
Let's suppose GTX380 has also 2,249X more transistors than GTX285.
So GTX380 = 3,15 billion transistors (instead of 3 million...) (1400*2,249)
With 2,249X scaling for transistor number, the 5870 (40nm) was only
1,284X bigger (die size) in relation with 4870 (55nm) (334mm2/260mm2).
So the hypothetical GTX380 (40nm) with 3,15 billion transistors will have also 1,284X scalling regarding die size.
So GTX380 (3,15 billion transistors) will be
603mm2. (470mm2*1,284)
So the real GTX380 with 3 billion transistors (with the logic that you are using) is going to be
less than 603mm2.
In the best case scenario will be 603mm2/3,15*3=575mm2.
But the logic thing is to be a little more than 575mm2, because the scaling in transistors (3-> 3.15) doesn't bring exact scaling in die density (example: 2,249 vs 1,284) (let's say bellow 591mm2)
The thing is that in 5870-4870 case the memory controller is the
same 256bit, while in the GTX380-GTX285 case the new memory controller has
smaller width than the old one. (GTX380=384bit, GTX285=512bit)
So you have to deduct a little bit die space from the new design (GTX380).
I guess (based on this logic) a good estimation is 576mm2 (same as GTX280)
The thing is that with a new architecture, new process technology and with the way that NV calculates the transistors (does the old numbers include cache transistors? does in this case the 3billion figure include cache transistors?) it is too difficult to say what the die size will be.