AVL tree size in terms of height
Let $n$ be the size of an AVL tree with height $h$. Then we have
$$ \log_2(n+1) \leq h \leq c\log_2(n+2) + b$$
where $c\approx 1.44$ and $b\approx -0.328$. So, 
$$ 0.8(1.6)^h\lt df^h=2^{-b/c}\left(2^{1/c}\right)^h=2^{(h-b)/c}-2 \leq n \leq 2^h -1 \lt 2^h $$
where $d\approx 0.85$ and $f\approx 1.618$.

Now you can see that the only number among 1.1, 1.9 and 2.1 that is between 1.6 and 2 is 1.9. Hence the answer is C.

I would not say it is asking for a tight bound.