Every instruction has 8 bit opcode.
Instruction size is 32 bit.
Immediate operand is 12 bit.
Some instructions have three register operands.
Instruction size - (3×(register operand)+opcode)=immediate operand
=> 32-(3n+8)=12
=>3n+8=20
=>3n=12
=>n=4
Now, the number of registers that the instruction set support is,
2^n =2^4 = 16
Therefore the correct option is (c) 16.
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