Endothermicity of Reaction
I believe that you would have to know at least these things to start with:


All enthalpies of bond dissociation are endothermic
Complete dissociation of triple bonds is more endothermic than double bonds, which is more endothermic than single bonds (when compared for the same elements).
The difference between going from a triple bond to a double bond is smaller than double to single, and that is smaller than going from single to complete dissociation (you referenced this in your question)


We can summmarize these facts with the following figure:



Here D(O-O) represents the bond dissociation enthalpy of an oxygen-oxygen single bond. The numbers inside of circles represent "steps" between each successively stronger bonds, and are there just for easier reference.

With that in mind, let's look at the options again, and count how many of each type of bond are broken and formed:

$ a) \space O_3(g) \rightarrow O_2(g) + O(g) $

Ozone has a resonance structure, which means that we need to consider the bonds as being somewhere between a single and double bond in strength. We could imagine them being about halfway up step 2.

Broken: 2 x single/double resonance bonds

Formed: 1 x double bond

So we are looking at an overall reaction that would be slightly more endothermic than the breaking of a single bond. To make that easier to see, think about taking these energetic "steps" using the figure above:

(+1.5, +1.5) - breaking the two resonance bonds

( -1, -2) - forming the double bond

$ b) \space  2O_3(g) \rightarrow 3O_2(g) $

Here we are breaking two sets of resonance bonds, but forming 3 double bonds. So our steps look like:

(+1.5, +1.5, +1.5, +1.5) - breaking six resonance bonds

(-1, -2, -1, -2, -1, -2) - forming three double bonds

There are three -1, -2 steps, each of which is larger than a +1.5 step. However, they are not much larger, and the total of the difference may or may not be greater than the fourth 1.5 step. This means we can't say for sure whether it is exo- or endo-thermic, but we can at least say it would be less endothermic than breaking a single bond. 

$ c) \space  3O_2(g) \rightarrow 2O_3(g) $

Now we are doing the opposite of (b). We still don't know the sign, but we know the magnitude is the same as (b), which was less than (a), so either way it can't be (b) or (c).

$ d) \space  O_2(g) \rightarrow 2O(g) $

Here we are breaking a double bond, and forming nothing. This is the clear winner so far, with steps +1, +2.

$ e) \space  O_3(g) + O(g) \rightarrow 2O_2(g) $

In the final case, we break two resonance bonds, and form two double bonds. Our steps are:

+1.5, +1.5, -1, -2, -1, -2

Since the formation of each double bond releases more energy than the dissociation of each resonance bond, this is clearly exothermic, as you stated in your question.

Taking all of this together, we have to choose (d) as the correct answer.