Hexadecimal — option (D).

What the question is after. Machine code is the pattern of bits the processor actually executes. Nobody writes or reads those patterns as bits, though, and the number system in which 8085 machine code is expressed — in every manual, listing, monitor program and hex keypad of the era — is hexadecimal.

Why hexadecimal and not something else. The 8085 has an 8-bit data word, and 8 bits split exactly into two groups of four. Each group of four bits is one hex digit, so every byte is precisely two hex characters — no padding, no ambiguity:

The correspondence is not merely convenient, it is exact: \(16=2^{4}\), so converting between binary and hex is a digit-by-digit substitution needing no arithmetic at all. Addresses tell the same story — the 16-bit address bus gives a range of \(0000_{H}\) to \(FFFF_{H}\), four hex digits covering all 64 KB.

Why octal fails is worth seeing, because it was the earlier convention: 8 bits do not divide into groups of three, so a byte needs 2⅓ octal digits and the leading digit carries only two bits. Octal suits 12-, 24- and 36-bit machines; hexadecimal suits 8-, 16- and 32-bit ones, which is why it displaced octal as byte-oriented processors took over.

The point of care. Strictly, the machine code is binary — that is what sits in memory and drives the instruction decoder. Hexadecimal is a human-readable representation of it. Examination questions of this form ask for the number system in which 8085 machine code is written and handled, and the expected answer is hexadecimal; the official key records (D).

BCD, for completeness, is not a machine-code notation at all. It encodes one decimal digit per nibble and exists for arithmetic on decimal quantities — which is why the 8085 provides DAA, the decimal adjust accumulator instruction, to correct a binary addition back into BCD form.

Hence, the answer is Hexadecimal.