#FileManagement

Question

consider a FAT system with 24 bits for each entry.For a 32GB disk with block size 32KB.What is the minimum size(in Mbits)of a file allocation in this system??

Answer 1

Disk = $2^{35}$ Bytes
Block Size = $2^{15} $ Bytes
Total possible number of blocks = $\frac{2^{35}}{2^{15}} = 2^{20}$ 1 million blocks
24 bits of 3 Bytes are required for 1 entry, for 1 million blocks 3MB space is required

So I think answer should be $2^{35} Bytes - 3*2^{20} Bytes $ OR
$2^{20}( 262,120) bits$ or 262,120 Mbits