machine has 2 byte = 16 bits instruction format, therefore, possible encodings = 2^16.

As the processor has 16 integer register, number of bits for one integer register = 4 (2^4 = 16)

As the processor has 64 floating point register, number of bits for one floating point register = 6 (2^6 = 64).

For type-1 category having 4 instructions each having 3 integer register operands (4*3 = 12 bits) will consume 4 * 2^12 = 2^14 encodings.

For type-2 category having 8 instructions each having 2 floating point register operands (2*6 = 12 bits) will consume 8 * 2^12 = 2^15 encodings.

For type-3 category having 14 instructions each having 1 integer register and 1 floating point register operands (4 + 6 = 10 bits) will consume 14 * 2^10 = 14336 encodings.

For type-4 category instructions, number of encodings left = 2^16 – 2^14 – 2^15 – 14336 = 2048.

For type-4 category having N instructions each having 1 floating point register operand (6 bits) will consume N* 2^6 = 2048 (calculated from previous step). Therefore, N = 32.