Your superscripts are all off by one with respect to the usual notation. That is, $\Sigma^0_n$ in the language of set theory means we have first-order quantifiers only, that is, over sets, whereas $\Sigma^1_n$ means we have $n$ alternations of second-order quantifiers (over classes).

If one is speaking of first-order formulas and set objects only, then the property in a model of set theory that every definable class $\{x\mid\varphi(x)\}$ has a definable element is equivalent to $V=\text{HOD}$, because the class of non-OD sets is certainly definable, but can have no definable element, and so it must be empty if the property is to hold; conversely, if $V=\text{HOD}$ holds, then there is a definable well-ordering of the universe, and so the definable class has a least member with respect to that well-order, and this object will be definable.

So the models of set theory in which every definable class has a definable element are exactly the models of V=HOD.

But you seem to be interested not in defining classes of sets, but rather in properties of classes $C$. That is, if $\varphi$ is a property of a class $C$ and there is a class $C$ such that $\varphi(C)$, then is there a definable class $C$ for which $\varphi(C)$?

Here, the answer is: not necessarily. For example, consider a model of NGBC that has a truth-predicate $T$ for first-order truth. For example, any model of Kelley-Morse set theory is like that, but considerably less than KM suffices. Now, the assertion that "$T$ is a truth predicate" is a first-order expressible property of $T$, since one need only say that $T$ obeys the Tarskian recursion. But there can be no definable class $T$ that is a truth predicate, because of Tarski's theorem on the non-definability of truth.

Meanwhile, there are models of NGB set theory that are pointwise definable, meaning that every set and class is definable without parameters, and in such a model, if there is a class $C$ with property $\varphi(C)$, then $C$ is already definable, because all classes in such a model are definable.

So the answer is that it is sometimes true, and sometimes not true, depending on your model of NGB.

Your superscripts are all off by one with respect to the usual notation. That is, $\Sigma^0_n$ in the language of set theory means we have first-order quantifiers only, that is, over sets, whereas $\Sigma^1_n$ means we have $n$ alternations of second-order quantifiers (over classes).

If one is speaking of first-order formulas and set objects only, then the property in a model of set theory that every definable class $\{x\mid\varphi(x)\}$ has a definable element is equivalent to $V=\text{HOD}$, because the class of non-OD sets is certainly definable, but can have no definable element, and so it must be empty if the property is to hold; conversely, if $V=\text{HOD}$ holds, then there is a definable well-ordering of the universe, and so the definable class has a least member with respect to that well-order, and this object will be definable.

So the models of set theory in which every definable class has a definable element are exactly the models of V=HOD.

But you seem to be interested not in defining classes of sets, but rather in properties of classes $C$. That is, if $\varphi$ is a property of a class $C$ and there is a class $C$ such that $\varphi(C)$, then is there a definable class $C$ for which $\varphi(C)$?

Here, the answer is: not necessarily. For example, consider a model of NGBC that has a truth-predicate $T$ for first-order truth. For example, any model of Kelley-Morse set theory is like that, but considerably less than KM suffices. Now, the assertion that "$T$ is a truth predicate" is a first-order expressible property of $T$, since one need only say that $T$ obeys the Tarskian recursion. But there can be no definable class $T$ that is a truth predicate, because of Tarski's theorem on the non-definability of truth.

Meanwhile, there are models of NGB set theory that are pointwise definable, meaning that every set and class is definable without parameters. For example, see my paper:

• Hamkins, Joel David; Linetsky, David; Reitz, Jonas, Pointwise definable models of set theory, J. Symb. Log. 78, No. 1, 139-156 (2013), http://dx.doi.org/10.2178/jsl.7801090. (ZBL1270.03101).

In such a model, if there is a class $C$ with property $\varphi(C)$, then $C$ is already definable, because all classes in such a model are definable.

So the answer is that it is sometimes true, and sometimes not true, depending on your model of NGB.