First, you haven't actually specified a particular field, since the field $F$ that you have will depend on your choice of $c$ and of $M$. For example, different nonstandard models can seriously affect even the cardinality of the field $F$ that you produce, so they are not all the same.
In fact, none of these fields is algebraic over $\mathbb{Q}$. To see this, let $a$ be any nonstandard integer in $M$ below the $\sqrt{c}$ in $M$. It follows that all the standard powers of $a$ are still bounded by $c$, even if you also multiply by rational numbers. So the $\mod c$ part of $F$ never arises when evaluating a polynomial over $\mathbb{Q}$ at $a$. So the only way such a polynomial can be zero at $a$ is if it is the zero polynomial, and so $a$ is transcendental over $\mathbb{Q}$.