No.
With a 6V supply the output of a 555 with no load is +4.7V. An emitter-follower will reduce it even more to about 4.0V or less.
I don't know if the speaker is fed DC and the current into an 8 ohm speaker is +500mA or if a coupling capacitor is feeding plus and minus 250mA.
No?
What do you mean, no? No, what?
You say that the output of a 555 with 6 v supply is 4.7V. The OP says he was getting 5V P-P. Don't argue with me about that; take it up with the OP. Maybe he was using the CMOS version of the 555.
And, as I said, 5v P-P is
more than enough to get a "sound" out of a speaker. If 5 v P-P is more than enough, 4 V P-P is still enough. After all, the requirement is "We were asked to use a 555 timer and any type of BJT amplifier to produce a sound out of the speaker."
Your own calculation indicates that "...the current is 6V/(470+100 ohms= 10.53mA which will be the peak current with no load." with the common emitter configuration the OP was trying out. That will be the maximum current in one direction with a load, and with RE of 100 ohms, he will only get ~60 mA in the other direction (with no speaker coupling cap).
Even if he only gets 250 mA with capacitor coupling and an emitter follower, that sure beats 10.53 mA one polarity and 60 mA the other polarity doesn't it?
He could bias the common emitter amp for
much higher quiescent current, but an emitter follower (with no output cap; a few hundred mA of DC in the speaker won't hurt it unless it is very small) is the obvious way to match the relatively high impedance of his 555 output to the low impedance of the speaker, when no additional voltage gain is needed, as is the case here. Maybe the lesson the instructor wants him to understand is that no voltage gain is needed here; it's an impedance matching problem.
And, with only a 6V supply, he can't possibly get more than 6V P-P with any configuration of BJT amplifier operating on that supply, and if he can get 4V P-P with the emitter follower, why bother trying to get more output voltage? There won't be much difference in SPL with 4V P-P compared with 6V P-P.
Also, the problem didn't say that the BJT amplifier has to be linear. The simplest solution would be to connect the speaker as the collector load, allow DC in the speaker, have no emitter resistor and use the BJT as a switch. Insert a suitable resistor between the 555 and the BJT base and possibly another resistor in series with the speaker if there is danger of overheating the voice coil.