Given , disk block address requires 32 bits = 4 Bytes and DB size is 1 KB = 210 Bytes .
If , The Disk Block Address = DBA bits
The Disk Block Size = DBSize Kbits
then maximum file size possible
= { n0 + n1 ( DBSize / DBA ) + n1 ( DBSize / DBA )2 + n3 ( DBSize / DBA)3 } * DB size
where n0 = direct disk block addresses,
n1= single indirect disk block addresses,
n2 = doubly indirect DBAs
n3 = triple indirect DBAs.
Given , n0 = 12 , n1 = 1 , n2 = 1 , n3 = 1
Now, DBSize / DBA = (210 / 4) = ( 210 / 22 ) = 28
Hence , Max file size possible = 12 + { (1 * 256 ) + (1 * 2562 )+ (1 * 2563 ) } * 1024 B
= ( 12 + 256 + 65536 + 16, 777 , 216 ) * 1024 Bytes
= 16, 843, 020 * 1024 Bytes
= 17, 247, 252, 480 Bytes
Reference :
https://gateoverflow.in/20557/elaborate-unix-inode