Here is a characterization of $\lambda$, but of course your question is somewhat ambiguous as to what counts as an answer.
For $n<\omega$, let $\alpha_n$ be the least ordinal $\alpha$ such that $\kappa<\alpha$ and $L_\kappa\preccurlyeq_n L_\alpha$. (This exists by weak compactness.) Then $\lambda=\sup_{n<\omega}\alpha_n$.
Proof: To see $\lambda\leq\sup_{n<\omega}\alpha_n$, let $T$ be a $\kappa$-tree which is $\Sigma_m$-definable from parameters over $L_\kappa$. Let $T'$ be the interpretation of such a definition of $T$, over $L_{\alpha_{m+3}}$. Then $T'\cap L_\kappa=T$, and $T'$ has nodes at level $\kappa$, and these give a $\kappa$-branch through $T$, hence in $L_{\alpha_{m+3}}\subseteq L_\lambda$.
To see the converse, fix $n$. Define the natural $\kappa$-tree over $L_\kappa$, searching for a wellfounded model $M$ such that $L_\kappa\preccurlyeq_n M$. This is definable over $L_\kappa$, at roughly $\Sigma_{n+3}$ or so (without using parameters). (At height $\eta<\kappa$ in the tree, where $\eta$ is a reasonably closed limit ordinal, we would have built the restriction $M\upharpoonright\eta$ of the planned eventual model $M$ on elements given by ordinals $<\eta$, and specified its $\Sigma_n$-theory on those elements, and would have also produced some isomorphism between $L_\eta$ and some $\in$-intitial segment of $M\upharpoonright\eta$ (and this must remain an $\in$-initial segment from this node onward). The $\Sigma_n$-theory in $L_\kappa$ of elements of $L_\eta$ must match that of their images in $M\upharpoonright\eta$. Whenever we specify an existential statement in the theory in one of these nodes (of height $\eta$), it must be witnessed in the same node with some element $<\eta$. (That is, if the node specifies the formula $\exists x\ \varphi(x,\vec{p})$, then there must be some $\alpha<\eta$ such that it also specifies $\varphi(\alpha,\vec{p})$.) We also demand that $M\upharpoonright\eta$ is wellfounded; any $\kappa$-branch will then give a wellfounded model, since $\kappa$ is inaccessible.)
Now just note that from a $\kappa$-branch through $T_n$, if $M$ is the corresponding model, then $M=L_\alpha$ for some $\alpha>\kappa$ such that $L_\kappa\preccurlyeq_n L_\alpha$. But we can't have such a branch $\in L_{\alpha_n}$, since we can't have such an $M\in L_{\alpha_n}$, since otherwise $L_{\alpha_n}$ could compute the Mostowski collapse of $M$ and that would give $\alpha<\alpha_n$, a contradiction. (But we get a branch through $T_n$ definable over $L_{\alpha_n}$, since there is a surjection $\kappa\to L_{\alpha_n}$ definable over $L_{\alpha_n}$.) So $\lambda\geq\sup_{n<\omega}\alpha_n$.
