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Given Maximum of 2K files...each file  in directory uses 32 bit(4byte).

so 2K Files use 8KB (in directory).

 

Now directory occupies 4 blocks, so 4 blocks = 8KB

1 block size= 2KB.

Each block is address with 16 bit,means there are total 2^16 blocks,

Hence size= 2KB*2^16= 128 MB
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Total number of addressable block are 2^16  ,  Total number of blocks needed for directory  = 4 , so maximum file size in blocks is 2^16  = 65536 .  Total memory for storing 2k file records of 32 bits each  =  2048*32/8 = 8192 bytes.  Total number of blocks need to store directory is 4 , so  4 * BlockSize = 8192   ,  BlockSize = 2048 bytes.  Total size of largest file in blocks(MB) = 65536*2048/2^20 =  128 MB
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“each block is addressed with 16 bits” means that there are 2^16 blocks 

solving for block size – 

there are 2k files means there will be 2K(2048)  entries in directory

each entry occupy 32 bits(4B) then 2K entries will occupy 2048 x 4B is directory size (8192)

directory occupy 4 blocks so size of each block = 8192/4 => 2048 bytes (2^11)

there are 2^16 such block so total size of memory is = 2^11 x 2^16 => 2^27 B or 128 MB

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