In the constructible universe, there is a definable well-ordering of the entire universe. The ordering, as I mention in [this MO answer](), is that $x$ preceeds $y$ when $x$ is constructed at an earlier stage of the $L$ hierarchy, or else they are constructed at the same stage, but $x$ is constructed at that stage by an earlier definition, or with the same definition, but with earlier parameters. 

If one only wants to well-order the real numbers, then one can extract from this definition a $\Delta^1_2$ ordering of the reals. The reason is that if $x$ is a real number of $L$, then it appears at some countable stage $L_\alpha$ for a countable ordinal $\alpha$, and the entire structure $L_\alpha$ is countable, and hence itself coded by a real. Furthermore, the $L$-order is absolute to any $L_\alpha$, since $L_\alpha$ knows about the $L_\beta$-heirarchy for $\beta<\alpha$. Also, if a countable structure is well-founded and thinks $V=L$, then it is $L_\alpha$ for some $\alpha$. 

Putting all this together, we get that the following are equivalent for any two reals $x$ and $y$:

 - $x\lt_L y$ in the $L$ order. 
 - There is some countable ordinal $\alpha$ such that $L_\alpha$ satisfies $x\lt_L y$.
 - There is a real $z$ coding a well-founded structure that thinks $V=L$ (and so this structure must be some $L_\alpha$) in which $x$ and $y$ are reals and the structure satisfies $x\lt_L y$. 
 - All reals $z$ coded well-founded structures $L_\alpha$ in which $x$ and $y$ are reals satisfy $x\lt_L y$. 

The third statement has complexity $\Sigma^1_2$, since being-well-founded is $\Pi^1_1$. Similarly the fourth statement has complexity $\Pi^1_2$, so overall the ordering is $\Delta^1_2.