The amount of ROM needed to implement a 4 bit multiplier is

Key idea: compute how many input combinations there are and how many bits are needed for each product.

• There are two 4-bit inputs, so total input bits = 4 + 4 = 8. That gives 2^8 = 256 distinct input combinations (or think of a 16 × 16 table = 256 entries).

• Each product can be up to 8 bits (15 × 15 = 225, which fits in 8 bits). So each entry must store 8 bits.

• Total ROM required = 256 entries × 8 bits = 2048 bits = 2 Kbits.

Therefore, the correct answer is 2 Kbits.