Emitter follower biasing with voltage divider
The book's approach is an estimation, on the assumption that the base current is negligible - which turns out to be not so great of an assumption here, as you found out.

your approach is more precise, but made the same mistake right here: 


  The maximum current through the divider without connected load is:
  Idiv=VccR1+R2=15280⋅103≈54μA


you didn't factor in the equivalent resistance on the base side: at 8.1v @ 10ua, that's equivalent to a 810K resistor (approximately Re * beta - see note below).

So the lower resistor R2 is paralleled by a 810K resistor. Once you factor that in your calculation, it will be alright.

Most people don't take that approach. for example, I typically set the current through R1/R2 to be 10x of the base current. That yielded R1 + R2 = 15v / 100ua = 150K. and go from there for R1/R2 individually. the 10x is picked to make sure that the base current is indeed negligible.

what it shows you is that 1) don't put too much stock in any book; and 2) don't take estimation too seriously. many times, good enough is indeed good enough.

edit:

note: for better approximation, some people would assume that the lower resistor is by-passed by a equivalent resistance of beta * Re -> 750K in this case, vs. 810K the real one calculated earlier. This approach works fairly well as an approximation.