Noise Figure calculation problem

I am recently reading a book, "CMOS Cellular receiver front-ends from specification to realization " by Johan Janssens。 In the subset 3.5.2, page41, there is a description as below:

?The required noise figure of the complete receive path is 8.87 dB, of which 3 dB has already been
assigned to the LNA. Taking into account the 18 dB gain of the LNA, the required noise figure
for the quadrature down-converter becomes 25.6 dB. However, in order to get some margin, the
noise figure specification of the receive path has been set to 5 dB, requiring a noise figure less
than 18.6 dB from the quadrature down-converter. For a single down-converter this boils down
to a required noise figure of 21.6 dB, provided that all the output noise between DC and 200 kHz
is taken into account. The detailed proof is given in Subsection A.3.2. ?

The system is of antenna,LNA, mixer cascoded.
But I calculate the required NF for mixer in the case of total NF of 8.87dB, but I cannot get the result of 25.6dB. What I get is a little more than 16dB. I just wonder what is wrong with my calculation formulation. The forms I calculated is Ftotal=Flna(1+s)+(Fssb,mixer -2)/A. The s is the image rejection, A is the gain of LNA, and Fssb,mixer is the noise factor in the case of single side band. I calculate the noise factor first, and then convert it into noise figure. I hope someone help me find my error. Thanks in advance.

In order to do the calculation you can use the Friis formula, that is:

NFtot = NFlna + (NFmixer - 1)/Glna

all expressed in linear, not in dB (of course linear = 10^(dB/10))

substituting your figures:

7.71 = 2 + (NFmixer - 1)/63.1

Solving with repect to NFmixer:

NFmixer = 5.71*63.1 + 1 = 361.301 linear units

NFmixer(dB) = 10*log10(361.301) = 25.6 dB

Thank you !