This is not a good exercise question, because none of the options is actually correct. It basically asks you to blindly apply a model which cannot yield the correct result.
Before continuing any deeper into the matter, let's have a look at the actual structure. I was not able to find a crystal structure, so I ran a quick optimisation on the RI-BP86/def2-SVP level of theory. For this purpose, this is accurate enough.
The bond angle $\angle(\ce{CNC}) = 144^\circ$ is right in between $120^\circ \dots 180^\circ$.
The (pure) VSEPR model must fail for this molecule, as it is only really working for central atoms with distinct and non-delocalised ligands.
For simple cases it performs very well in describing the approximate shape of a molecule. Once you have gotten this down, you can refine predictions with other tools. Please note, the lone pairs take up more space argument is actually a post-rationalisation and should not be used within the framework of VSEPR theory. More reliable predictions can be made with Bent's rule, which says that more electronegative ligands will concentrate p-orbital contributions towards them, decreasing the bond angle (this is a very, very brief summary).
As far as the VSEPR model goes, one should concentrate on simple coordination patterns:
- linear; the bond angles will be close to $180^\circ$
- trigonal; the bond angles will be close to $120^\circ$
- tetrahedral; the bond angles will be close to $109^\circ$
- higher coordinations like trigonal bipyramidal, octahedral, etc.
(TL;DR) From that we see that the central nitrogen has three ligands, so the general coordination around it is trigonal and the bond angle should be close to $120^\circ$. This is the answer, which your book will accept as correct, because it blindly applies the VSEPR model.
Applying a bit of Bent's rule won't help much. Carbon is slightly less electronegative than nitrogen, hence one would expect a slight opening up of the angle (more s-character directed towards carbon), but that's not enough to push it to $144^\circ$.
More importantly, the $\ce{-NCO}$ group is highly delocalised. Another very important concept comes into play: resonance. Ignoring the underlying structure for a moment, these are the most important contributors:
$$\ce{Me-N=C=O <-> Me-N^+\bond{3}C-O^-}\tag1\label{eq:resonance}$$
Within the concept of resonance, the positions of the nuclei will not change, and the structures do not exist on its own. A single structure of a set is therefore always an incomplete description. Please read What is resonance, and are resonance structures real?
The left structure in \eqref{eq:resonance} is basically what we have described above with VSEPR. From the depiction on the right, we would expect a $\angle(\ce{CNC}$ of about $180^\circ$, because triple bonds are coordinated linear. Since these are electronic descriptions of the same structure, we can expect the actual (underlying) structure to be somewhere in between these hypothetical forms.
The observed molecular structure is always the result of a delicate balance of effects. So repulsion of nuclei, electronic repulsion, electronic attraction, dispersion, etc. will all have an effect. Many of these interact quite complicated with each other. Sometimes our used models will yield good approximations, sometimes they don't.
In this particular case, we see that VSEPR alone fails; including resonance will refine the prediction. However, without a more rigorous approach, it is impossible to predict the bond angle correctly, i.e. we do not know the contribution of the right structure without doing some more maths. I used a quantum chemical approach, which is obviously nothing you could do in an exam. This is one of the reasons why I dislike this question.
A quick word on hybridisation:
The concept of hybridisation is independent of VSEPR. Many books will throw them together, but it is actually incorrect. You can make a prediction of the shape of a molecule with VSEPR. After that, you can use the molecular structure to choose the closest matching hybridisation model (if any at all). As such, hybridisation is always a result from a particular structure, never a cause. Therefore, your initial assertion is reverse:
My current assumption is that the $\ce{e-}$ in nitrogen in $\ce{CH3NCO}$ forms sp2 hybridization (and therefore should forms a shape like ethene).
Correct is, because the structure around nitrogen is trigonal, the electronic structure can be approximated with sp2 orbitals.
Why $\angle\ce{CNC} \approx 104^\circ$ cannot be correct
In order for this angle to be achieved, the nitrogen first needs to be surrounded by four ligands (or electron pairs), giving us an angle of about $109^\circ$ to start with. Then more p-character needs to go into the bonds, reducing the angle. Hence the ligands must be more electronegative than the other ligands. A popular example is dichloromethane, or even water (but that is a bit more complicated). See the Wikipedia article on Bent's rule, our Q&A Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?, and multiple other questions with regard to Bent's rule.
