If an algorithm has time complexity $T(n) \in \Theta (\log_2 n)$, then it means that $T(n)$ is bounded above and below by constant factors of $\log_2 n$ for sufficiently large $n$. This definition actually leaves a lot of room for possible $T(n)$, and the question in your textbook makes some additional assumptions (or uses a very loose definition of "approximate").
If $T(n) = \log_2 n + c$, then $T(2n) = T(n) + 1$, so $T(10^6) = 20$ implies $T(2 \cdot 10^6) = 21$.
However, it could also be that $T(n) = k \log_2 n + c$, in which case $T(2n) = T(n) + k$. Note that the complexities $\Theta(\log_2 n)$ and $\Theta(\log n)$ are equivalent, and you would typically choose the simplest possible function to express the growth rate. By writing $\Theta(\log_2 n)$, the textbook is abusing notation to let you know that $k = 1$.
Even then, more complicated expressions are possible e.g. $T(n) = \log_2 n + A \log_2 (\log_2 n) + B \cos n$. Only for sufficiently large $n$ do those additional terms become irrelevant. For any fixed $n$, the answer to your question could be arbitrarily skewed. Therefore, I would say the question is formally meaningless, and whatever intuition it helps build is outweighed by the confusion it creates.