Answer:
electronic processing device
Answer:
2 bytes
Explanation:
Size of the MAR ( memory address register ) = 18 bits = 2 bytes
Given that
address space = 256 K = 2^8
number of address location= 2^8 * 2^10 ( where 1K = 2^10 )
= 2^18
Given the number of address location = 2^18 ;
Hence 18 bits are required to store the address of the instruction
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