For me, the most foolproof way to identify the endo and exo products is to look at the stereochemistry in the product. Consider first a standard intermolecular Diels–Alder reaction:
I labelled the substituents on the diene $\mathrm{R^t}$ and $\mathrm{R^c}$, for trans and cis respectively, to describe their position with respect to the single bond in the middle of the diene.
What we can see from this is that the endo product has the electron-withdrawing group (EWG) placed cis to $\mathrm{R^t}$ (and trans to $\mathrm{R^c}$). (OK, sorry, the naming is a bit confusing; but I'm sure you can make sense of it.) The opposite would be true of the exo product: in that case, the EWG would be cis to $\mathrm{R^c}$ (and trans to $\mathrm{R^t}$).
The intramolecular case is harder to draw a transition state for (user55119 has done it nicely!), but this doesn't change the conclusion drawn above. In your case, the diene only has a single substituent which is of the $\mathrm{R^t}$ type (the substituted double bond is trans configured). This substituent ends up becoming the five-membered ring:
It is quite clear from the final product that the EWG is trans to the $\mathrm{R^t}$ substituent. So, in agreement with user55119's analysis, the product shown is actually exo.