The consistency strength of the failure of $\square(\kappa)$ for non-weakly Mahlo inaccessible cardinal $\kappa$ (in particular, for the first strongly inaccessible) is higher than weakly compact. For example it implies that $0^\#$ exists.
http://www.jstor.org/stable/27590333
We know that global square holds in $L$, i.e. there is a sequence $\langle C_\alpha | \alpha \text{ is singular limit ordinal} \rangle$ with the properties: $\forall \alpha,\, \text{otp } C_\alpha < \alpha$ and for every accumulation point $\beta$ of $C_\alpha$, $C_\alpha \cap \beta = C_\beta$.
In $V$, there is a club $D\subset \kappa$ through the singular cardinals below $\kappa$. If $0^\#$ doesn't exists, the covering theorem will imply that those cardinals are singular in $L$ as well, and therefore $C_\alpha$ is defined for every $\alpha \in D$.
Now we can define a $\square(\kappa)$ sequence, $\langle E_\alpha | \alpha < \kappa \rangle$ in $V$ as follow:
- For $\alpha \in \text{acc } D$, $E_\alpha = C_\alpha \cap D$ (unless this set is bounded below $\alpha$. In this case, $\text{cf }\alpha = \omega$ and we can choose $E_\alpha$ to be arbitrary $\omega$ sequence cofinal in $\alpha$).
- For $\alpha \notin \text{acc }D$, $E_\alpha = \alpha \setminus (\max D\cap \alpha)$
It's easy to see that $\langle E_\alpha | \alpha < \kappa\rangle$ is coherent (since the global square is coherent) and thread-less (since every thread meets $D$ club many times, but $\text{otp }E_\alpha < \alpha$ for every $\alpha \in D$).
Edit: As Mohammad noted, we can get better lower bound by using more appropriate inner model. We need to violate global square on the singular cardinals. By $\square$ On the singular cardinals (Theorem 1), we need at least many measurable cardinals of uncountable Mitchell order in order to get a model without global square on the singular cardinals.
