Yea, i thought so. But taking 4096 Bytes as the block size, it gives the max file size for triple indirection to be around 262 Bytes. Now doesn't it make it too impractical as it amounts to a file size of 4 EXABYTES?
How 262 Bytes?
Shouldn't it be 211*212=8MB? (i.e. # of disk block addresses*size of 1 block)
Ohh sorry. I didn't copy the complete question it seems. Have now edited it!
For triple indirection, it amounts to 262 .