$20$ should be the correct answer.
To find the number of digits(not bits) in a number, say $x$, we need to find the ceiling of log base ten of $x$, that is $\left \lceil \log_{10}x \right \rceil$.
Given $\log{2} = 0.30103$, it seems a bit incomplete as the base of logarithm is not mentioned.
However since $10^{0.30103} \approx 2$, so it can be inferred that in $\log{2} = 0.30103$ the base of log should be $10$.
Coming to the original question, the number of digits in $2^{64}$ would be $\left \lceil \log_{10}\left ( 2^{64} \right ) \right \rceil = \left \lceil 64\log_{10}2 \right \rceil = \left \lceil 64 \times 0.30103 \right \rceil = \left \lceil 19.26592 \right \rceil = 20.$
