Consider a file system using the indexed allocation. The block size is 512 bytes and the pointer to a block takes 2 bytes. The largest file size that a one level indexed allocation method could handle is KB. Recall that 1KB is equal to 210 bytes. Answer -1 if there is no limit based on the provided information.

128 KB      

Explanation:

Given data :

The block size = 512 bytes

Block pointer size = 2 bytes

In an indexed allocation, a special block named Index Block contains all the pointers to the blocks that the file had occupied.

Therefore, the maximum number of block pointer in 1 block :

= 256

Thus, a block can hold maximum 256 pointers.

It is given the level of indexing is one. So,

Maximum size of the file =  maximum number of block pointers in 1 file x block size

                                          = 256 x 512

                                          = 128 KB   (∵ 1 KB = )

Therefore, the largest file that 1 level if indexed allocation method can handle is 128 KB.