You really should specify the edition (there are 3 of them) and the exercise numbers. Page numbers would also help.
To put this into context, you are discussing the following exercises found on page 66 of the 3rd edition:
\$\quad\quad\quad\quad\$
I don't know where you got the right answers for either, above. I just have to accept them.
You have a circuit like this:
simulate this circuit– Schematic created using CircuitLab
If you refer to Figure 1.100(A) on page 49, you will see this:
I've circled their selection (as you quote it.)
That chart does not take into account the loading by the \$10\:\text{k}\Omega\$ load. But it does show that the values (assuming an unloaded RC filter) look very, very close to what you quoted.
One thing to keep in mind is that for simple RC circuits, there is an inflection point in the phase angle curve right at \$45^\circ\$. And this also corresponds to where the output is \$-3.01\:\text{dB}\$ below its maximum.
These two ideas are often conflated. One might say \$-3.01\:\text{dB}\$ when what is really meant is the phase angle inflection point. Your low-pass case is an example of where the phase angle inflection point does not occur where the output is \$-3.01\:\text{dB}\$ relative to the input source. However, it is still true that the output, at the phase angle inflection point, will be \$-3.01\:\text{dB}\$ relative to where it would be near DC. It's important to get your head wrapped around what I just wrote in this paragraph.
In the answer you say is right, the critical angular frequency (the point where the phase shift is \$45^\circ\$) will be \$\omega_{_0}=\frac1{\left(R_1\,\mid\mid\, R_{_\text{L}}\right)\,C_1}=72500\:\frac{\text{rad}}{\text{s}}\$ (about \$f_{_0}\approx 11.539\:\text{kHz}\$) and the near-DC gain will be \$A=\frac1{1+\frac{R_1}{R_{_\text{L}}}}\approx 0.8621\$ (about \$-1.29\:\text{dB}\$.) However, at \$\omega_{_0}\$ the output will be down another \$-3.01\:\text{dB}\$, for a total of about \$-4.30\:\text{dB}\$. So this is a case where the \$-3.01\:\text{dB}\$ point for the output occurs before reaching the critical angular frequency. The reason is that you have a resistor divider that also attenuates the input signal, even near DC.
Just for validation:
So their question isn't entirely clear on that point. But at least you can see from Figure 1.100(A) where they might have gotten their values. The effective source impedance is enough lower (less impacted by) than the load impedance to be acceptable, perhaps. (It could be made still lower, of course.)
Can you provide the source for the answers they gave you?