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. (A Lowenheim-Skolem argument shows that $F$ can be found as you describe of any desired infinite cardinality cardinality.)
But to answer your question, none of these fields is algebraic algebraic over $\mathbb{Q}$. To see see this, let $a$ be any nonstandard nonstandard integer in $M$ whose finite powers are bounded below below $c$ in $M$ (see below). It follows that any polynomial polynomial over the positive integers$\mathbb{N}$ evaluated at $a$ is still less than than $c$ in $M$. So the $\mod c$ part of the field operations of $F$ never arises never arise when evaluating a polynomial over $\mathbb{N}$ atover $a$. So the only way such two such polynomials can agree$\mathbb{N}$ at $a$ is if they are the same. Thus, nothe problem reduces to showing that if $p$ is a nontrivial polynomial over $\mathbb{Z}$ at, then $p(a)\neq 0$ for nonstandard $a$ can bein $0$ except for$M$, and this follows because the zero polynomialbasic eventually-unbounded asymptotic behavior of such polynomials is provable in your theory. Thus, and so $a$ is transcendental over $\mathbb{Q}$ in your field $F$.
Edit. Finally, here is a quick-and-dirty way to see that there is such a nonstandard element $a$, whose finite powers are bounded by $c$ in $M$. Let $a$ be the nearest nonstandard integer to $c^{1/N}$, where $N=\sqrt{\log c}$ as interpreted discretely in $M$. Since $N$ is nonstandard, it follows that the finite powers of $a$ are below $c$, and since $\log a\equiv\frac 1N\log c$, it follows by the choice of $N$ that $a$ is nonstandard. But I expect that there is an easier method.