in Operating System
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consider a system where each file is associated with a 16 bit number.for each file ,each user should have the read and write capability.how much memory is needed to store each user's access data?

A) 16KB                B) 32KB                C) 64KB              D) 128KB

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is the ans A 16KB

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each file is associated with a 16 bit number.  this means that no. of files $= 2^{16}$ and $16bit$ are required for read access and $16-bits$ for write access.

Am i correct @sanjay sir, and @Kapilp

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Each file requires 2 bits (read and write)

Associated bits with each file are 00,01,10,11 .
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means no. of files $= 2^{16}$

1-bit for reading, 1 for writing, which add to total $2-bits$

memory needed $= 2^{16}*\frac{2}{8} = 16KB$
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