= Solution
For radial free fall, let $E=f\dot t_A>0$ be Alice's conserved <Killing energy>. The inward branch of the <timelike geodesic> has $\dot r=-\sqrt{E^2-f}$, hence
$$
\dot u_A=\dot t_A-\frac{\dot r}{f}=\frac{E+\sqrt{E^2-f}}{f}\sim\frac{2E}{f}.
$$
With $\delta=r-2M$, the near-horizon <geodesic equations> give
$$
f\sim\frac{\delta}{2M},\quad \delta\sim E(\tau_H-\tau),\quad t_A=-2M\log(\tau_H-\tau)+O(1),\quad r_* =2M\log(\tau_H-\tau)+O(1).
$$
The constants in the logarithms are understood to make their arguments dimensionless. Therefore $u_A=t_A-r_*=2t_A+O(1)$. For an outgoing signal reaching the fixed-radius Bob, $t_B=u_A+r_*(r_B)$, so $t_B=2t_A+O(1)$. The <redshift> consequently behaves as
$$
\frac{\lambda_B}{\lambda_A}\propto e^{t_B/(4M)},\qquad \boxed{M=\frac1{4a}}.
$$
This is the answer when $t$ is Bob's <Schwarzschild time>, as appropriate to the stated observation. It is the <Schwarzschild surface gravity> $a=\kappa=1/(4M)$. If $t$ instead means the emission <Schwarzschild time> $t_A$, the same <redshift> is proportional to $e^{t_A/(2M)}$ and gives $M=1/(2a)$. The two answers use different clocks, not different dynamics. If the measured exponential uses Bob's <proper time> $\tau_B$, its rate is $1/(4M\sqrt{f_B})$. In SI units the reception-time result is $M=c^3/(4Ga)$.
The reception-time exponent also holds for smooth radial infall crossing the future <Schwarzschild event horizon> with finite nonzero $dr/d\tau$: regular <ingoing Eddington-Finkelstein coordinates> give finite $v$ there, $u=v-2r_*\sim-4M\log(\tau_H-\tau)$. Thus no special value of $E$ is needed.
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