Fundamental Minimum Noise Floor
General theory taken as reference point

Shannon-Hartley theorem indicates that with sufficiently advanced coding techniques transmission at channel capacity can occur with arbitrarily small error.


As S/N increases, the information rate can increase while still
preventing errors due to noise; an infinite information rate is
possible if SNR tends to infinite (no noise), irrespective of
bandwidth;
As B increases, similarly the information rate may go up, but with B going to infinite, channel capacity is limited to \$1.44 S/\eta\$.


The power spectral noise density is
$$\eta=k_{0}T_{eq}$$
and Teq is the equivalent noise temperature (as noted in a comment above) and not the medium temperature.

If we are sending binary digits (as appears seeing data-rate without other details on phases and levels) and we assume the symbols associated to the two digits (0 and 1) have the same power, then the average energy per bit is given by signal power S multiplied by data rate R:
$$E_{b}=SR$$

It is possible to find a limit for the energy per bit: the noise power is $$N=\eta B,$$ the symbol rate is taken equal to the channel capacity $$R=C$$
Thus
$$\dfrac{E_{b}}{\eta}=\dfrac{B}{C}\left(2^{C/B}-1\right)$$

This equation is plotted for variable B/C values, finding the limit of -1.59dB (Shannon limit), in this case for a rate that has reached channel capacity (R=C).



The region labeled as "cannot be implemented" is the one where partial transmission is possible (see David K's question), but not the entire message with a reducing error.

Now let's consider the attainable spectral efficiency for a given SNR.
Spectral efficiency rho is used to relate the data rate and the occupied bandwidth W (at most the whole channel bandwidth B):
$$R=\rho B$$
it is possible to rewrite the Shannon limit as
$$SNR&gt;2^{\rho}-1$$
Linking back to the ratio of energy per bit (signal bit, not information bit) and noise power spectral density,
$$Eb/\eta = SNR/\rho$$,
we have
$$\dfrac{E_{b}}{\eta}&gt;\dfrac{2^{\rho}-1}{\rho}$$

With rho indicating different levels of exploitation of channel bandwidth, we obtain different minimum signal-to-noise ratios: 
$$\dfrac{E_{b}}{\eta}\geq\begin{cases}
\begin{array}{ll}
1.76\mathrm{dB} &amp; \rho=2\\
0\mathrm{dB} &amp; \rho=1\\
-1.59\mathrm{dB} &amp; \rho\rightarrow0
\end{array}\end{cases}$$

Note. For small SNR (power-limited channels) the attainable spectral efficiency increases linearly with the SNR, whereas for large SNR (bandwidth-limited channel) the increase is only logarithmic.

The -1.59dB indicates that it is possible for lower data rates to use negative SNR, although not so low.

Interpretation of data in question

1) The term "frequency" of 1 Hz and 106204 Hz refers to the data rate R or the bandwidth B? for what is said below it ought to be the bandwidth.

2) The expression "in maximum 1 hour" is ambiguous: if it is the duration of the transmission of 64 bits, we shall assume exactly 1 hour, that  determines the data rate as 64/3600 bit/s, and the term "frequency" shall be the bandwidth.

3) The term "noise floor" should come directly from B*k0*Teq (or B*k0*Tamb, assuming a noise figure of 1); with k0 = -228.6 [dBW/K/Hz], summing 10*log10(Tamb) and 10*log10(B) we have: B=1 Hz => -203.98 dBW, B=102604 Hz => B=-153.71 dBW. And the latter matches with the comment @Tony Stewart.

Check of first eq in the question

Data rate R = 64/3600 = 0.0178

Bandwidth B = 1 Hz or 106204 Hz

Rho = 0.0178 or 3.35 10^(-7)

SNRmin = 2^rho - 1 => -19.1 or -66.3 dB