Horowitz's exercise 2.2
Wow, I have the first edition of this book - I guess I got it 25 years ago ;-).
One of the approaches I "learned" from this bookset (=including workbook) is that you should either build it to try or simulate it and try to understand what you see from there.
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Without referring to the book, here is my approach.

At 50mA, the minimum DC gain is rated at 60 (not 30).  I'll use 30 anyway to stick to your analysis.
We allow a voltage change of 5% which is 0.25V.  B-E voltage drop has a big impact on this.
If the quiescent collector current is 10mA, the voltage drop can be 0.1V @30mA.

So we have about 0.15V left for the load change versus the bias resistors.
The load looks about 30 times bigger (no need to be too precise), lets call it Re.
First we suppose that R3 is part of the source impedance (Rs).
Vout/Vopen = (Re/(Re+Rs).  Hence for a drop of .15V (3%) we have \$0.97 = 6k/(6k+R_s)\$.  \$R_s=6000/.97-6000=180\Omega\$.
15V is about 3 times 5V so \$R_1\$ is about two times \$R_2\$ and \$R_s\$ is about \$\frac{1}{1/2*R2+1/R2} = \frac{1}{1.5/R2} = R_2/1.5 = 185\Omega\$.

So R2=277.5Ohm, which we round to 270 - a value we can find in the manufactured values.
We need 5V+0.6V on the base, so $$R1 = \frac{15V-5.6V}{5.6V} * 270 = 453 Ohm$$.  The closest is 430 Ohm (keeping in mind that the voltage will drop because of R3).
So bias voltage is $$15V*\frac{270}{430+270}= 5.78V$$.  With 470 Ohms, it would be 5.47V, which is too low.

If I set 10mA as quiescent collector current, then I have R3=500Ohm, or 510 Ohm for a standard value.
The effective R2 is then 270 Ohm//(510 Ohm*30)= 265 Ohm. The bias voltage is 15*265/(265+430)= 5.72V and I get about 5.1V at the output.

Adding the load of 200Ohm, the effective R2 is 265 Ohm//6kOhm=253.8 Ohm.
The bias voltage is \$15V\frac{253.8}{253.8+430}=5.57V\$.  Our B-E voltage drops itself from -0.6V to -0.7V, the output is at 4,87V.
The total drop is 5.1-4.87=0,23V, just under 5%.

Another check of the drop is to check how much the voltage change in R1 is due to the change in current.  The current change is 25mA/30=0.83mA.  Hence the voltage drop is 430*.00083=0.38
One could think that the extra current of 25mA results in an extra base current of 0.83mA and hence a drop over the 430 Ohm resistor of 0.36V, higher than the 0.23V.
However, with a drop of 0.13V, the current in the 265 Ohm equivalent resistor drops by 0.5mA.  Hence the current in R1 increases by only 0.33mA which corresponds to a drop of 0.14V (roundings, etc. will explain the difference of 0.1V).
So this check using currents also verifies.

Therefore, R1=430, R2=270 and R3=510Ohm looks like a good solution.
One could tune R1 and R2 a bit to find an output value closer to 5V when there is no load, but 5.1V within reasonable limits of the design goal (2%)
The estimated voltage drop is just under 5%.

In this process, I've used only the standard resistor values.  I did not do a very precise calculation, but followed a 'hands-on' approach because practical components add variation anyway.

That's where the Art is: knowing how to get to a result without resorting to the full mathematical formula which gets you a 0.001% precise solution.

Also, once you get a solution, you have to verify it.  I did it by checking the actual result without and with a load.
I kept in mind the variation of the B-E voltage by checking the datasheet while already choosing a reasonable quiescent current (10mA looked reasonable).
In the endd, it all adds up to a good solution.

You can notice that my solution does not respect your constraints - R2>184!!

Now this does not indicate where your error is, but it does indicate that the difference is not an order of magnitude.

Your second approach says that R1*(Vlow/Vdrop)=Rload/Rout.  There is an issue with units where.  On the left I have a resistor value, on the right a ratio.  I can't explain this formula.
It is not clear how you get to the inequality of the first version either.
It looks probable that your mistakes are there.