Define $$\hat w_\lambda = \arg\min_w L(\Theta, X, y) + \lambda \|w\|_2^2.$$ We know that $\lim_{\lambda \to \infty} \hat w_\lambda = 0$, due to the penalty $w \mapsto \|w\|_2^2$ having the origin as its minimizer.
Sycorax points out that, similarly, $\lim_{\lambda \to \infty} \left\{ \arg\min_w L(\Theta, X, y) + \lambda \|w-c\|_2^2 \right\} = c.$ This successful generalization may lead us to propose the estimator $$\tilde w_\lambda = \arg\min_w L(\Theta, X, y) + \lambda \mathrm{pen}(w),$$ where $\mathrm{pen}$ is a function whose minimizer satisfies some property that we seek. Indeed, Sycorax takes $\mathrm{pen}(w) = g(\|w\|_2^2 - 5)$, where $g$ is (uniquely) minimized at the origin, and, in particular, $g \in \{|\cdot|, \, (\cdot)^2\}$. Therefore $\lim_{\lambda \to \infty} \|\tilde w_\lambda \|_2^2 = 5$, as desired. Unfortunately, though, both of the choices of $g$ lead to penalties which are nonconvex, leading the estimator to be difficult to compute.
The above analysis seems to be the best solution (maybe up to the choice of $g$, for which I have no better one to suggest) if we insist on $\lambda \to \infty$ as being the unique interpretation of "tends to" described in the question. However, assuming that $\|\arg\min_w L(\Theta, X, y) \|_2^2 \geq 5$, there exists some $\Lambda$ so that the minimizer $\hat w_\Lambda$ of OP's problem satsifes $\|\hat w_\Lambda\|_2^2 = 5$. Therefore $$\lim_{\lambda \to \Lambda} \left\| \hat w_\lambda \right\|_2^2 = 5,$$ without needing to change the objective function. If no such $\Lambda$ exists, then the problem of computing $\arg\min_{w : \|w\|_2^2 = 5} L(\Theta, X, y)$ is intrinsically difficult. Indeed, there's no need to consider any estimator besides $\hat w_\lambda$ when trying to encourage natural properties of $\|\hat w_\lambda\|_2^2$.
(To enforce that a penalized estimator attains a value of the penalty which is not achieved by the unpenalized estimator seems highly unnatural to me. If anyone is aware of any places where this is in fact desired, please do comment!)