The A_V, Q, and f_p are:

A_V=\frac{30 \ k\Omega }{51 \ k\Omega }+1=1.59
Q=\frac{1 }{3-A_V }=\frac{1}{3-1.59} =0.709
f_p=\frac{1}{2\pi RC}=\frac{1}{2\pi(47 \ kΩ)(330 \ pF)}=10.3 \ kHz

It takes a Q of 0.77 to produce a ripple of 0.1 dB. Therefore, a Q of 0.709 produces a ripple of less than 0.003 dB. For all practical purposes, the calculated Q of 0.709 means that we have a Butterworth response to a very close approximation. The cutoff frequency of a Butterworth filter is equal to the pole frequency of 10.3 kHz. Notice in Fig. 19-32b that the pole frequency is at approximately 1 dB. This value is 3 dB down from the passband gain of 4 dB.