Let $G$ be a finite group. Then the group algebra $\mathbb{Q}G$ trivially contains $\mathbb{Q}$.
But when (i.e. for which $G$) does the augmentation ideal $I_G=\{\sum_{g\in G} r_g\,g \mid \sum_{g\in G} r_g=0,\ g \in G,\ r_g \in \mathbb{Q}\}$ of $\mathbb{Q}G$ contain a subring isomorphic to $\mathbb{Q}$.
I know that this is true for $\mathbb{Q}S_3$ with the subring $\{\frac{2}{3}r\ 1_G-\frac{1}{3}r\ t-\frac{1}{3}r\ t^2\mid t^3=1_G,r\in\mathbb{Q}\}$.
Always.
Since $\Bbb QG$ is a semisimple $\Bbb Q$ algebra, $\Bbb QG=I_G\oplus J$ for another ideal $J$, and as rings $I_G$ and $J$ are both $\Bbb Q$ algebras, each having an idempotent acting as a multiplicative identity.
You just take the idempotent $e\in I_G$ which is the identity, and $e\Bbb Q\cong \Bbb Q$.
That is exactly what happened in your example, btw, since $\frac{2}{3}\ 1_G-\frac{1}{3}\ t-\frac{1}{3}\ t^2$ is the idempotent generator of $I_G$ in that ring.
Of course, you can make the exact same case for any semisimple group algebra $F[G]$ that the augmentation ideal contains a copy of $F$ and all of its subfields.
I couldn't immediately draw a conclusion about all algebras with nontrivial radicals, but there do exist counterexamples there. For example, $F_2[C_2]$ for the cyclic group of order 2 $C_2$ has a nilpotent augmentation ideal, so it cannot contain a copy of $F_2$.