Answer:
31 bits.
Explanation:
Given, total number of registers = 55
Total instructions = 60
Size of memory = 16 KB
Now, no of registers are 55. We find the next greater or equal power of 2 which is 64 = 26. Hence, 6 bits are required to represent a register operand.
Number of instructions = 60. We find the next greater or equal power of 2 which is 64 = 26. Hence, 6 bits are required to represent a instruction.
Size of memory = 64 KB = 26 * 210 * 23 bits = 219 bits. Hence, 19 bits are required to represent a memory location.
Now, an instruction has 2 parts, opcode and operand. As given there are only two address instructions which are memory operand and register operand.
Hence, total bits would be: 6 bits (opcode) + 6 bits (register operand) + 19 (memory operand) = 31 bits.
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