Final answer:
The maximum possible file size in this file system with 2048-byte blocks and 32-bit disk block pointers can be calculated by summing the sizes represented by the direct, singly indirect, double indirect, and triple indirect pointers. The result is approximately 17.59 terabytes (TB).
Step-by-step explanation:
In this file system, the maximum possible file size can be determined by calculating the sum of the sizes of the direct, singly indirect, double indirect, and triple indirect pointers. The direct pointers represent the number of direct blocks that can be accessed without indirection, and since each block is 2048 bytes in size and there are 12 direct pointers, the total size represented by the direct pointers is 2048 * 12 = 24576 bytes.
The singly indirect pointer allows access to a block of 2048-byte pointers, each of which can point to a data block. So, the size represented by the singly indirect pointer is 2048 * 2048 = 4194304 bytes.
The double indirect pointer allows access to a block of 2048-byte pointers, each of which points to a block of 2048-byte pointers that in turn point to data blocks. Hence, the size represented by the double indirect pointer is 2048 * 2048 * 2048 = 8589934592 bytes.
Similarly, the triple indirect pointer allows access to a block of 2048-byte pointers, each of which points to a block of pointers which in turn points to a block of pointers that point to data blocks. Therefore, the size represented by the triple indirect pointer is 2048 * 2048 * 2048 * 2048 = 17592186044416 bytes.
To obtain the maximum possible file size, we add the sizes represented by the direct, singly indirect, double indirect, and triple indirect pointers: 24576 + 4194304 + 8589934592 + 17592186044416 = 17592190630988 bytes. This is equivalent to approximately 17.59 terabytes (TB).