ROM is Read-Only Memory. We can just read from it, not write to it. Hence, all the possible results need to be already stored in the ROM.
A 4-bit multiplier multiplies two 4-bit numbers.
There can be $2^{4}$ such numbers.
So, total possible multiplication pairs could be $2^{4} * 2^{4}=2^{8}$
The result would be max: $(1111)_{2}*(1111)_{2}$
$=> (15)_{10}*(15)_{10}$
$=> (225)_{10}$
And 225 requires 8 bits.
Hence, we potentially need 8 bits for the $2^{8}$ possible combinations.
$=> 2^{8} * 8 bits$
$=> 2^{11} bits$
So, Option D
Edit: Alternatively,
We know that multiplying an $n$ bit number with and $m$ bit number results in a number which is of at most $(m+n)$ bits.
Hence, maximum result size: $8$ bits.
Total combinations: $2^8$
Memory required: $2^{11} bits$
PS: Addition of two $n$ bit numbers results in a max $n+1$ bit number