Let's call bit strings that don't contain two consecutive 0's "such bit strings".
Such bit strings of length 0 = 1. // $\epsilon$
Such bit strings of length 1 = 2 // 0 and 1
Such bit strings of length 2 = 3 // 01, 10 and 11.
Such bit strings of length 3 = 5 //010, 011, 101, 110, 111
Such bit strings of length 4 = 8 // 0101, 0110, 0111, 1010, 1011, 1101, 1110, 1111
We observe that at length k, "such bit strings" = sum of the previous two cases.
So, Such bit strings of length 5 = 8 + 5 = 13
Such bit strings of length 6 = 13 + 8 = 21. (Answer)
This is given by a(n) = a(n-1) + a(n-2).
This is actually the Fibonacci sequence.
The number of bit strings that do NOT contain two consecutive 1's is also the same.