Memory size $=512\text{ KB}=2^{19}$ bytes $\Rightarrow$ address field needs $19$ bits.
Registers $=6 \Rightarrow$ bits per register $=\lceil \log_2 6\rceil=3$ bits. For two registers $\Rightarrow 2\times3=6$ bits.
Total operand bits $=19+6=25$.
Opcode bits $=32-25=7 \Rightarrow$ maximum distinct instructions $=2^{7}= {128}$.
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