If a computer has eight-byte addressability and needs three bits to access a location in memory, what is the total size of memory in bytes?

64 byte of memory

Explanation:

The computer has 8 bytes addressability and the memory only requires 3 bits to access its entire memory size. Remember that a computer is an electronic device and it stores data and receives instructions in binary of 1 and 0. So for an eight-byte addressable computer,

total memory size = 2^(3 bits) x 8 (computer's addressability)

= 8 x 8

= 64 bytes of memory.

i think the answer is a or d

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