Correct Option: D

The normal size of ROM is n × 2ⁿ
∴ Now, we are multiplying two n-bit numbers. So, the resultant has 2n bit.
Hence, the size of the ROM is 2n × 2²ⁿ
In the question, n = 4
Hence, ⇒ 2 × 4 × 2^{2 × 4}
⇒ 8 × 2⁸ ⇒ 2³ × 2⁸
⇒ 2 × 2¹⁰ ⇒ 2 k bit
OR
As there is two independent 4 bit inputs, so you need 2 × 4 = 8 address lines A 4 bit × 4 bit multiplication has 8 bit result, you need 8 data lines.
∴ Rom needs to be at least 256 × 8 = 2 k bit.