The Galois cohomology group $H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z})$ classifies cyclic cubic extensions $K/\mathbb{Q}$ (specifically: the non-trivial elements correspond to Galois cubic field extensions $K/\mathbb{Q}$ together with a choice of isomorphism $\mathrm{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/3\mathbb{Z}$).

Let $k = \mathbb{Q}(\mu_3)$. There are restriction and corestriction maps
$$\mathrm{Res}: H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}) \to H^1(k, \mathbb{Z}/3\mathbb{Z}), \quad \mathrm{Cores}: H^1(k, \mathbb{Z}/3\mathbb{Z}) \to H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}).$$ 
Restriction followed by corestriction is multiplication by $2$ on $H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z})$. As each element is $3$-torsion, it follows that $\mathrm{Res}$ is an isomorphism and that $\mathrm{Cores}$ is surjective (possibly it is even an isomorphism?).

But as $\mu_3 \subset k$, it follows from Kummer theory that 
$$H^1(k, \mathbb{Z}/3\mathbb{Z}) \cong H^1(k, \mu_3) \cong k^*/k^{*3}.$$
Composing with corestriction, we therefore obtain a surjective map
$$f: k^{*}/k^{*3} \to H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}).$$
>> Can the map $f$ be made explicit? Namely, given a non-cube $a \in k^*$, what is the cyclic cubic extension of $\mathbb{Q}$ induced by $f$?

I know that the corestriction $H^1(k, \mu_3) \cong k^*/k^{3*} \to \mathbb{Q}^*/\mathbb{Q}^{*3} \cong H^1(\mathbb{Q}, \mu_3)$ is just usual norm map. But this doesn't seem to help here.