So if I understand your question correctly, you are looking for examples in which $\kappa$ is weakly inaccessible, but not strongly, yet satisfies $\kappa^{<\kappa}=\kappa$.
Conceptually, your example is the easiest. First note that to find such an example you need to start with a situiation which has consistency strength at least a (strongly) inaccessible, as ZFC+ a weakly inaccessible has the same strength as ZFC + a (strongly) inaccessible (this is as any weakly inaccessible is strongly inaccessible in Gödels constructible universe).
Now in this situation, there must be some cardinal $\lambda<\kappa$ so that $2^\lambda=\kappa$. Moreover, after $\lambda$ up to $\kappa$, the continuum function must constantly evaluate to $\kappa$. That is, there is some minimal $\lambda_0<\kappa$ so that for all $\lambda_0\leq\lambda<\kappa$, $2^\lambda=\kappa$.
So the only thing that really is variable is the value of $\lambda_0$. The minimal possible value is, of course, $\omega$, which is exactly your example. Starting with an inaccessible $\kappa$ + GCH, for any regular $\delta<\kappa$ we can find a forcing extension in which $\lambda_0$ is exatly $\delta$.
I am quite sure that it is even consistent that $\lambda_0$ is singular, though this would require much higher consistency strength than just an inaccessible (but dont quote me on that...) [edit: upon further reflection, you may quote me on that. It is however quite interesting how such a singular $\lambda_0$ can look. A prominent result of Shelah gives $\lambda_0\neq\aleph_\omega$. In fact his PCF theory proves that a singular $\lambda_0$ must be a fixpoint of the $\aleph$-function, if I am not mistaken.]. So this might be an interesting example.
Here is another example: In the usual construction of a model in which $2^\omega$ is real-valued measurable (or equivalently, in which there is a countably additive measure on the reals, that measures every subset of $\mathbb R$ and which extends the Lebesgue measure) one starts with a measurable cardinal $\kappa$ (so in particular, $\kappa$ is (strongly) inaccessible) and adds with a specific forcing $\kappa$ new reals. In the extension, $\kappa=2^\omega$ satisfies $\kappa^{<\kappa}=\kappa$, is weakly inaccessible and real-valued measurable. So this is a construction where such a situation naturally pops up.
Best Answer
Well, it's easier to do it by contrapositive.
Since $\kappa$ is strongly inaccessible, the set of strong limit cardinals is a club below $\kappa$. If the regular cardinals form a stationary set, the intersection with the aforementioned club is stationary. So there is a stationary set of inaccessible cardinals.
In particular, there are unboundedly many inaccessible cardinals below $\kappa$, so it has to be the $\kappa$th inaccessible.