In Koepke's paper, "Turing Computations On Ordinals", one has the following (well-known) result:

A set $x$ is ordinal computable from a finite set of ordinal parameters if an only if it is an element of the constructible universe.

In Sacks' survey article, "E-Recursive Intuitions", one finds these (possibly related) results:

Proposition 3.1. $V$=$L$ iff $L$ is E-recursively enumerable... Proposition 2.3. There exists a $\Delta_1$ definable class that is not E-recursively enumerable.

Now, since the above theorem of Koepke's characterizes the constructible sets as those ordinal computable from a finite set of ordinal parameters (note that the *pure sets* are sets of ordinals), one should be able to characterize the axiom $V$=$L$ as

'Every set of ordinals is ordinal computable from a finite set of ordinal parameters.'

Substituting this characterization for $V$=$L$ in Sacks' theorem, one has

'Every set of ordinals is ordinal computable from a finite set of ordinal parameters iff $L$ is E-recursively enumerable.'

It is interesting to note that Sacks' proof of Theorem 3.1 from his paper gives an indication of what must happen if $V$$\neq$$L$ (most set theorists do not believe that $V$=$L$ is an 'acceptable' axiom):

Proof of 3.1. Suppose $\forall$$x$($x$$\in$$L$ $\leftrightarrow${$e$}($x$)$\downarrow$) and $V$$\neq$$L$. Then for some $b$$\notin$$L$, {$e$}($b$)$\uparrow$. By Proposition 2.2 ["If $A$ is E-recursively enumerable, then $A$ is $\Delta_1$ definable."--my comment], divergence [$\uparrow$--my comment] is $\Sigma_1$ definable. Then by Levy-Shoenfield absoluteness, {$e$}($x$)$\uparrow$ for some $x$$\in$$L$.

Hence the question asked in the title.

Three other questions:

Can one 'force' $L$ to be not E-recursively enumerable?

If $V$$\neq$$L$, are there constructible, non-E-recursively enumerable sets?

Since $L$ is usually considered a 'class', is E-recursion defined for classes (since it is based on Normann's "Set recursion")? (Hope this is not too silly a question....)