Solution

The correct answer is option2

Solution:

Size of the disk block = 2048 Bytes

disk address = 32 bits = 4 Bytes

Number of addresses per block = 2048/4 = 512 = 2⁹

Maximum file size = 12 Direct Blocks + 1 indirect block + 1 double indirect block

                            = 12 + 2⁹ + 2⁹ * 2⁹

                            = 12 + 2⁹ + 2¹⁸

The size of each block = 2¹¹ Bytes

Maximum size of the file = ( 12 + 2⁹ + 2¹⁸ ) × 2¹¹ Bytes

∴ Maximum size of the file =   513.0234 ≈ 513 MBytes