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

1 K b i t s

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64 b i t s

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128 b i t s

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2 K b i t s

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Solution

The correct option is D $2 K b i t s$
$4 - b i t \times 4 - b i t \Rightarrow 8 - b i t$
$R O M s i z e = 2^{(n + n)} \times (n + n)$
$R O M s i z e = 2^{4} \times 2^{4} \times 8 b i t s = 2 K b i t s$

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