I need to find all number fields with absolute value of discriminant $\le 20$.
Using Minkovsky's theorem I understood that it should be quadric or cubic extension. The case of quadric is very easy. As for cubic case I understood that it should have 2 complex embeddings and his ideal class group should be trivial. But I don't know to how find all of them.
I'm going to assume you mean that the absolute value of the discriminant has to be $\le 20$. You seem happy with the quadratic case, and have correctly narrowed down the cubic fields to complex cubic fields.
Hint. The trick here is that the smallest value the discriminant of a cubic field can take (in absolute value) is $23$, namely $\mathbb{Q}[x]/(x^3-x^2+1)$ whose discriminant is $-23$, so there are in fact no cubic fields of discriminant $|\Delta|\le 20$.
As the OP has rightfully seen, it remains to exclude the possibilities $\Delta_K=-15,-16,-19,-20$ for a cubic number field $K$.
We can use class field theory to do this. For example, the cases $-15$ and $-19$ are excluded since they are squarefree and $\mathbb{Q}(\sqrt{-15})$ and $\mathbb{Q}(\sqrt{-19})$ have class numbers not divisible by three (they are $2$ and $1$ respectively). For the details see my answer on this question
Hilbert class field of a quadratic field whose class number is 3
A small modification of the proof in the link also settles the case $-16$: if $\Delta_K=-16$ occurs for a cubic number field $K$ then only $2$ ramifies in $K$ and is in fact wildly ramified, since total ramification would result in $\mathrm{ord}_2(\Delta_K)=2$. But then by the same argument we see that the normal closure $N$ of $K$ yields an unramified Abelian extension of $\mathbb{Q}(\sqrt{-16})=\mathbb{Q}(i)$ of degree $3$, which cannot happen as $\mathbb{Q}(i)$ has class number $1$.
Finally, suppose that $\Delta_K=-20$ for a cubic number field $K$ and let $N$ be its normal closure. As with the previous arguments, since $\mathrm{ord}_5(\Delta_K)=1$ we have non-total ramification of $5$ in $K$, so that $N/\mathbb{Q}(\sqrt{-5})$ is unramified over the prime of $\mathbb{Q}(\sqrt{-5})$ over $5$. Now if $2$ is also non-totally ramified (i.e. wildly ramified) we again see that $N/\mathbb{Q}(\sqrt{-5})$ is an unramified Abelian extension, which cannot happen as $\mathbb{Q}(\sqrt{-5})$ has class number $2$. If $2$ is totally ramified in $K$ then we see that the unique prime $\mathfrak{p}=(2,1+\sqrt{-5})$ in $\mathbb{Q}(\sqrt{-5})$ lying over $2$ ramifies in $N$, and since this ramification is tame (and also the only ramification) we would get that the Abelian extension $N/\mathbb{Q}(\sqrt{-5})$ has conductor $\mathfrak{p}$, so that $N$ would be contained in the ray class field $H_{\mathfrak{p}}$ of $\mathfrak{p}$. However we have the exact sequence $$ 0 \to (\mathcal{O}/\mathfrak{p})^{*}/\mathrm{im}[\mathcal{O}^{*}]\to Cl_{\mathfrak{p}}\to Cl_{\mathbb{Q}(\sqrt{-5})}\to 0, $$ for the ray class group $Cl_{\mathfrak{p}}$, which implies that $[H_{\mathfrak{p}}:\mathbb{Q}(\sqrt{-5})]=2$, which is a contradiction as $[N:\mathbb{Q}(\sqrt{-5})]=3$.
