 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 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 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.