Correct Answer - Option 2 : 35 KBytes
The correct answer is option 2
DATA:
In the index node of a Unix- style file system,
Direct Block pointer = 8
Single Indirect Block pointer = 1
Double Indirect Block pointer = 1
Disk block address = disk block entries size = 8 B
Disk Block size = 128 B
CALCULATION:
Number of entries in a block = \(\frac{disk\; block\; size}{disk\; block\; address\; size} = \frac{128\; B}{8\;B} \) = 24
File size = (8 direct + 1 singel indirect + 1 double indirect) × Block size
File size = (8 × 1 + 1 × 24 + 1 × 24 × 24 ) × 128 B
File size = (210 B + 211 B + 215 B )
File size = (1 + 2 + 32)KB
∴ The maximum size of file is 35 KB or 35 KBytes