Let $L$ be the field of algebraic real numbers over $\mathbb{Q}$. Let $K$ be a non-trivial algebraic field extension over $L$ in the field of complex numbers. Does it follows that $K$ coincides with the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$?
In the famous math problem source "Problems and solutions in mathematics" published by World Scientific, we can see the following problem.
Problem 1411: (a) Suppose you are given a field $L$, $\mathbb{Q} \subseteq L \subseteq \mathbb{C}$, such that $L/\mathbb{Q}$ is algebraic and every finite field extension of $K/L$, $K \subseteq \mathbb{C}$ is of even degree. Show that every finite field extension of $L$ must in fact have degree equal to a power of $2$. (b) Show that such a field $L$ actually exists.
Then question 1411(a) can be easily resolved by considering arbitrary finite Galois extension of $L$ and applying Sylow theorem to make use of Sylow subgruop of degree power of $2$ of the Galois group. For the question (b), the author of the book solved with the choice $L$ to be the field of all algebraic real numbers to conclude that every finite algebraic extension over $L$ is of degree power of $2$.
My question is that is there any algebraic extension of $L$ other than itself and the algebraic closure of $\mathbb{Q}$.
PS: It would be nice if the field $L$ with the property in 1411(a) is unique.