Let $F$ be a quartic extension of $\mathbb{Q}$. Let $\mathbb{Q}_{k} = \mathbb{Q}(e^{\frac{2\pi i}{k}})$.
Let $\mathbb{Q}_d$ be the maximal Cyclotomic field contained in $F$ (note that $\mathbb{Q}_{d_1}, \mathbb{Q}_{d_2} \subseteq F \implies \mathbb{Q}_{lcm(d_1,d_2)} \subseteq F$, and that if a primitive root of unity of order $k$ is contained in $F$ we have $\mathbb{Q}_{k} \subseteq F$. Thus, $\mathbb{Q}_{d}$ is well-defined and contains all the roots of unity contained in $F$).
Note that $\mathbb{Q}_{d} =\mathbb{Q}_{2d}$ when $d$ is odd, so we can assume the either $d$ is odd or divisible by $4$.
We have the following tower of extensions: $\mathbb{Q} \subseteq \mathbb{Q}_{d} \subseteq F$. This implies that $\varphi(d) = [\mathbb{Q}_{d} : \mathbb{Q}] | [F:\mathbb{Q}] = 4$. The relation $\varphi(d) | 4$ implies, together with the assumption on $d$, that $d | 5$ or $d|12$ or $d|8$.
So, the distinct options:
$d=5, 8$ or $12$, in which case $\phi(d)=4$ and $F=\mathbb{Q}_{d}$.
$d =3$ or $d=4$, in which case $\phi(d)=2$ and $F$ is a quadratic extension of $\mathbb{Q}_{d}$: $F=\mathbb{Q}(i,\sqrt{a+bi})$ or $F=\mathbb{Q}(\omega,\sqrt{a+b\omega})$
$d=1$, in which case $F$ doesn't contain roots of unity other than $\pm 1$.