Thickness of Depletion region, transistor
No, you're incorrect.

The Doping doping levels are:

E: highest doping level

B: less than E

C: weakest doping, less than B

It needs to be like that because the ratio of B/E doping sets the beta

The collector has the weakest doping level so that the depletion layer will be large so that it can handle a large reverse voltage. Remember that  in the active region the BC junction is reverse biased !

The thickness of the base must be small, the base must be thin because the carriers coming from the emitter should be have already passed through the base and enter the collector region before they realize:

Oh, oops ! Too late to recombine in the base.

And, Oh, blast now we're pulled to the collector contact (because that will have a large positive voltage, for an NPN that is).

That is what makes the base current small and forces the carriers to go to the collector instead of recombining in the base.

The depletion region in the base will be smaller than the depletion region in the collector because the base has a higher doping level ! Your assumption that the base has a lighter doping than the collector is not true. It is the collector that has the lightest doping.