I know that $Q(i)$ is the only one that fulfills the statement. But I'm not entirely sure how to get there.
I would define an isomorphism $g$ from $\mathbb{Q(i)}$ to $K$ using the identity when restricted to $\mathbb{Q}$.
As $\mathbb{Q}(i)=\{a+bi:a,b \in \mathbb{Q}\}$, then $K=g(\mathbb{Q}(i))=\{a+bg(i):a,b \in \mathbb{Q}\}$.
On the other hand, $i^2 + 1 =0$, which implies that $(g(i))^2+1=0$, hence $g(i)=\pm{i}$. That means that if $g(i)=-i$, then $g(-i)=i$. And if $g(i)=i$, then $g(-i)=-i$.
How does the last part imply that there is only one isomorphsim? My initial thought was that there were two.
If $K \subseteq \mathbb C$ is isomorphic to $\mathbb Q(i)$ then $[K:\mathbb Q]=2$. Moreover, there is $\alpha \in K$, such that $\alpha^2=-1$. So $\alpha = \pm i$ and $\mathbb Q(i) \subseteq K$. Since $[\mathbb Q(i):\mathbb Q]=2$, we must have $K=\mathbb Q(i)$.