Correct : b

The processor uses a 16-bit (2-byte) instruction format with 16 registers, so each register field needs 4 bits. Variable-sized opcodes mean unused opcode space at one level can be extended to the next level by adding more bits.
First, figure out how many bits each instruction type leaves for the opcode:
M-type has 2 register fields (4+4 = 8 bits) and a 6-bit immediate, totalling 14 bits of operands. This leaves 16 − 14 = 2 bits for the opcode. So M-type opcodes are 2 bits wide, giving 2² = 4 possible M-type opcodes. Only 2 are used, leaving 2 unused 2-bit prefixes.
R-type has 3 register fields (4+4+4 = 12 bits) of operands, leaving 16 − 12 = 4 bits for the opcode. R-type opcodes are 4 bits wide. To avoid conflicting with M-type''s 2-bit opcodes (00 and 01), R-type must use the 2 remaining 2-bit prefixes (say 10 and 11) and extend each to 4 bits. This gives 2 × 2² = 8 possible R-type opcodes. We need 7, so both unused M-type prefixes are consumed for R-type, and only 1 out of 8 R-type slots remains unused.
C-type has 1 register field (4 bits) and a 6-bit offset, totalling 10 bits of operands, leaving 16 − 10 = 6 bits for the opcode. C-type opcodes are 6 bits wide. The only space left is the 1 unused 4-bit R-type prefix, which can be extended by 2 more bits to form 6-bit C-type opcodes. This gives 1 × 2² = 4 possible C-type opcodes.
Correct answer: B — 4 ✓