A processor has 64 general-purpose registers and 50 distinct instruction…

Answer: 20 bits

Explanation:

• Total instruction length = 32 bits.

• Opcode bits = ceil(log2(50)) = 6 bits (since 2^5 = 32 < 50 ≤ 64 = 2^6).

• Register bits = ceil(log2(64)) = 6 bits for each register identifier. This instruction uses one explicit register operand (the destination/source register), so use 6 bits.

• Immediate bits = 32 - opcode bits - register bits = 32 - 6 - 6 = 20 bits.

Note: If an ISA required separate source and destination register fields for this instruction, that would consume more register bits and reduce the maximum immediate size. For the given assembly form which specifies a single register and an immediate, 20 bits is the maximum immediate field size.