Correct : b

Explanation:
To find the maximum number of bits available for an immediate operand, we need to calculate how many bits of the 32-bit instruction must be allocated to the Opcode (instruction type) and the register fields.
1. Calculate Opcode Bits:
The processor supports 50 distinct instruction types. To uniquely represent 50 operations, the number of bits required for the opcode field is:
    2⁵ < 50 ≤ 2⁶ → 6 bits are required.

2. Calculate Register Bits:
The processor has 64 general-purpose registers. To address any one of the 64 unique registers, the number of bits required per register field is:
    2⁶ = 64 → 6 bits are required per register.

3. Analyze the Target Instruction Type:
Look at the example given: ADD R1, #25 (which represents $R1 = R1 + 25$).
• This is an instruction that involves one destination register ($R1$), one source register (also $R1$ since it reads and modifies it), and one immediate operand ($25$).
• Therefore, the instruction structure must allocate bits for: [ Opcode ] [ Register Destination ] [ Immediate Value ].

4. Calculate the Remaining Bits for the Immediate Operand:
Subtract the bits consumed by the opcode and the single referenced register field from the total instruction size:
    Bits for Immediate Operand = Total Bits - Opcode Bits - Register Bits
    Bits for Immediate Operand = 32 - 6 - 6 = 20 bits.

5. Conclusion:
The maximum size of the immediate field for this instruction type is exactly 20 bits, matching option (b).