Answer
$6\text{ bytes}$
Work Step by Step
Total instruction size = Op code size+ Address-1 size+ Address-2 size
=> Total instruction size = 6+18+18 =42 bits=42/8 bytes =5.25 bytes
So 6 bytes will be required for one instruction.
Invitation to Computer Science 8th Edition
by Gersting, Judith L.; Schneider, G. Michael
Published by
Cengage Learning
ISBN 10:
1337561916
ISBN 13:
978-1-33756-191-4
Chapter 5 - Exercises - Page 274: 19c
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