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Answer must be d that is 2.018 gb

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No of address in a block = $2^{11}/2 = 1024$

maximum file size

= $8 * 2KB$ + $1 * 1024 * 2KB$ (1 single indirect block will have 1024 address, each will point to a 2KB block) + $1 *1024*1024*2KB$

= $16KB + 2MB + 2GB$

= $2.018 GB$ (Approximately)

Similar question was asked in BARC 2020, they asked for exact answer. (it was of the order of MB not GB)

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