Each i-node has a total 15 block addresses which means 15 bit block addresses.
So each i-node size = (2^15)B.
Now let’s see how many i-nodes are required to address the entire disk space, i.e.512 GB
= Total Size / I-node Size
= 512 GB/ (2^15)B
= (2^39)/(2^15)
= 2^24
So ‘24’ bits are required to address the entire disk space which is equal to 3 Bytes.
That means Block Entry Size= 3 Bytes
Now to support file size up to 1 GB, we need to calculate the number of i-node or blocks required,
= File Size/ I-node Size
= 1 GB/ (2^15)B
= (2^30)/(2^15)
=2^15
Now to address 2^15 block we need a space = No_of_blocks * Block_Entry_Size = (2^15)*3 Bytes
Now question says to find out this space in units of block addresses, So just divide by 1 block size.
= (2^15)*3 / (2^15) = 3 = 3 block addresses
