Let $X$ be any integral scheme, and let $\eta \in X$ be the generic point. (In particular, this includes any irreducible variety over a field, regardless of whether it's smooth or projective.) Then the residue field $\kappa(\eta)$ at $\eta$ is the function field $K(X)$ of $X$, that is, the field of all rational functions on $X$. This is also equal to the stalk $\mathcal{O}_{X, \eta}$, which is already a field (no need to take the field of fractions).
This is proven in Stacks Project, Lemma 29.49.5 (tag 01RV), though a few of the steps in that proof are unnecessary when there's only one irreducible component. The correspondence between elements of $\mathcal{O}_{X, \eta}$ and rational functions on $X$ essentially amounts to the fact that every nonempty open subset of $X$ contains the generic point $\eta$, which follows from irreducibility. Since $X$ is also reduced, every nonzero rational function on $X$ is invertible on some nonempty open subset, so $\mathcal{O}_{X, \eta}$ is already a field and hence is isomorphic to the residue field $\kappa(\eta)$.
When $X$ is an irreducible variety over a field $k$, the function field of $X$ is an extension of $k$ of transcendence degree equal to the dimension of $X$. This follows from the Noether normalization lemma. In particular, the degree is infinite except in the zero-dimensional case where $X = \operatorname{Spec}(L)$ for some finite extension $L/K$.