Exhaustive search times: 2 to power k = 100 hours - double k, how many hours

Question

An exhaustive search (i.e. checking all combinations of values) takes 100 hours to go through all permutations where a binary key has a length of k.

$2^k$ = 100 hours where k is the number of digits in the binary key and 2 is the base (0 or 1).

If the length of the key is increased to 2k. How long will it take?

i.e. $2^{(2k)}$ = ? hours

Answer 1

$2^{2k} = 2^k \cdot 2^k$, so your answer in the comments is right.