= Solution
A <register file> with 256 entries needs $\log_2 256=8$ <bits> per register index. Under the intended assumption that eight times as many instructions are obtained by expanding the primary <opcode> field, its width grows from six to $6+\log_2 8=9$ <bits>. Thus \b[<opcode>: 9 <bits>; rs, rt and rd: 8 <bits> each].
These four fields alone require $9+3(8)=33$ <bits>, already exceeding a 32-<bit> <instruction encoding>. Retaining the five-<bit> shift amount and six-<bit> function field of a <MIPS R-type instruction> gives \b[44 <bits>]. The processor's data word need not become wider merely because it has more registers.
The <opcode> conclusion is conditional: MIPS also uses secondary function fields, so an eightfold increase in the number of semantic operations does not by itself force a nine-<bit> primary <opcode>. That is the straightforward field-expansion interpretation of this exercise.
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