If you multiply the page size with the block size you will end up with bit², which is not a common storage unit, so that can't be right. There is also no 1 gigabyte device available.
I find that datasheet rather hard to understand. In their figure 2, they list the device organization like the following:
- 1 Page = (4k + 128)Bytes
- 1 Block = 64 Pages = (256k + 8k)Bytes
- 1 Device = 8192 Blocks = 16.896 Mbits (why the hell the sudden bits?
-> 2GB, the smallest device)
Now I don't know where the number of blocks is coming from, as well as I don't know where the number of pages is coming from. So I'd assume they are constant.
Based on that:
We have a page size of 4kB and a block size of 256kB (which seems like a redundant information if each block is 64 pages), so that matches with the above calculation to a 2GB device (using 8192 blocks per device).
Furthermore your calculation of number of planes times size of planes seems also reasonable
110_b turns to 4 gigabit planes, and
10_b to 4 planes, so 16 gigabit, which is also 2GB.
So I think it is safe to assume that this is a 2GB device.