To an experienced person, it is obvious at a glance that the transistor will be in heavy saturation.
I am not sure exactly where you went off track. But I do notice that in your question, you state two things that are mutually exclusive. You state that VB = VC, and you also state that VB=0.746 and VC = 0.024. That is an obvious fallacy in your question.
Conceptually, an NPN configured this way (which is very common) goes into saturation as the base current increases. Roughly speaking (assuming base supply and collector supply are the same), when the RB/RC = beta, the transistor is somewhere near the beginning of saturation. As RB gets smaller, it goes more and more into saturation.
In this case, RB/RC = 1, so you know the transistor is in very heavy saturation. You can also see this by the very low drop from collector to emitter, which is another hallmark of saturation.
When a transistor is used as a saturated switch, the ratio of Ic to Ib is lower than the datasheet beta. The ratio of the actual currents is sometimes called the "forced beta." Typical values of "forced beta" are 10 to 20. When the base and collector drive are the same voltage, you can just use a 10:1 rule where the base resistor is 10x the collector resistor.
When the system is not that simple, you can roughly calculate forced beta by assuming realistic values for Vbe and Vce and estimating the currents.
I am trying to present you with an intuitive and practical framework rather than a textbook approach. Hopefully it will help.