The method I'll describe is called Cyclotron Resonance, and it's a neat way to directly measure $m^*$ by using a fixed magnetic field $\boldsymbol B$.
The equation of motion of the electrons in a certain material, when in presence of a magnetic field $\boldsymbol B$ are
$$
m^*\dot{\boldsymbol v}=-e\boldsymbol v\times \boldsymbol B -\frac{m^*}{\tau}\boldsymbol v\tag{1}
$$
where $\tau$ is the relaxation time$^1$ of the electrons (in general, $\tau^{-1}$ is a very small number, so for now we might take $\tau^{-1}=0$; it will be important later). If we take $\tau^{-1}=0$, then the solution of $(1)$ is well-known: the electron moves in a circular orbit, with angular frequency
$$
\omega_c=\frac{eB}{m^*} \tag{2}
$$
By measuring $\omega_c$ for different values of $B$ we can get a very precise measurement of $m^*$. But, the obvious question, how can we effectively measure $\omega_c$ in a laboratory? The answer is surprisingly easy, as we'll see in a moment.
If, in the situation above, we turn on a monochromatic source of light (say, a laser) with frequency $\omega$, there will be an electric field $\boldsymbol E\;\mathrm e^{-i\omega t}$, and the new equations of motion will be
$$
m^*\dot{\boldsymbol v}=-e(\boldsymbol E(t)+\boldsymbol v\times \boldsymbol B) -\frac{m^*}{\tau}\boldsymbol v\tag{3}
$$
By using the ansatz $\boldsymbol v(t)=\boldsymbol v_0\;\mathrm e^{-i\omega t}$, and solving for $\boldsymbol v_0$ (left as an exercise), you can easily check that in this case, $\boldsymbol v(t)$ will be proportional to $\boldsymbol E(t)$ (which should be more or less intuitive). For example, if we take $\boldsymbol B$ in the $z$ direction, then $\boldsymbol v$ is given by
$$
\boldsymbol v_0=\frac{e}{m^*}\begin{pmatrix} i\omega-1/\tau&\omega_c&0\\-\omega_c&i\omega-1/\tau&0\\0&0&i\omega-1/\tau\end{pmatrix}^{-1}\boldsymbol E \tag{4}
$$
where $^{-1}$ means matrix inverse.
This system will absorb energy from the source, so that the transmitted light will be less intense than the incoming light. The absorbed power is just $\text{Re}[\boldsymbol j\cdot\boldsymbol E]$, and as $\boldsymbol j\propto \boldsymbol v$, it's easy to check that
$$
P\propto \text{Re}\left[\frac{1-i\omega \tau}{(1-i\omega\tau)^2+\omega_c^2\tau^2}\right]\propto \frac{1}{(1-\omega^2\tau^2+\omega_c^2\tau^2)^2+4\omega^2\tau^2} \tag{5}
$$
Now, if we vary $\omega$, the power $P$ changes, and from $(5)$ we can see that there will be resonance when $(\omega^2+\omega_c^2)\tau^2=1$. In practice, $\omega\tau\ll 1$, so the resonant frequency is $\omega\approx\omega_c$:
$\hspace{100pt}$
where the lines correspond to $\tau=0.1,\;0.5,\;1,\;3$, from green to blue. As you can see, for $\tau\to 0$, the resonance tends to $\omega_c$, so by measuring the resonant frequency we get the value of $\omega_c$, i.e., the value of $m^*$.
$^1$ the relaxation time $\tau$ is related to the mean free path: $\ell\sim v\tau$. Taking $\tau^{-1}\approx 0$ means that we assume the electron performs many cyclotron orbits before colliding with anything (ions, impurities,...).
Best Answer
Dear John, note that 23.85 Å is equal to 2.385 nm, while the observed 4.3 nm is approximately two times larger.
There is a simple error in your calculation that exactly fixes the factor of two. Note that the actual calculation you should have done has a radius proportional to $1/m$ and the correct $m$ that you should substitute is the reduced mass of the two-body problem governing the relative position of the two particles.
The reduced mass is $m_1 m_2 / (m_1+m_2)$.
Now, the important point is that an exciton is not a bound state of the effective electron and a superheavy nucleus: instead, it is a bound state of an effective electron and an effective hole - a larger counterpart of the positronium (an electron-positron bound state).
Assuming that both the electron and hole masses are equal, 0.26 $m_0$, the reduced (and still also effective) mass is 0.26/2 $m_0$ = 0.13 $m_0$, and the resulting $a$ is twice as big as your result, 4.77 nm - assuming that your arithmetics is right.
The deviation from 4.3 nm is not too large but I can only handwave if I were trying to pinpoint the most important source of the discrepancy. It could be a different effective mass of the hole; finite-size effects caused by the fact that the silicon atoms were not quite uniformly distributed inside the exciton, and so on.
Update
Oh, in fact, I noticed that your properties table does include a special figure of the effective hole's mass and it differs from the electron mass: 0.38 $m_0$. So the reduced mass is $$\frac{0.38\times 0.26}{0.38+0.26} m_0 = \frac{0.0988}{0.64} m_0 = 0.154 m_0 $$ and the calculated radius is $$ \frac{11.7}{0.154} \times 0.53\ Å = 40.3\ Å. $$ Well, this is 7 percent too small, much like the previous one was 7 percent too big. ;-)
Hydrogen atoms with composite heavy fermions
Concerning your second question, as you clearly realize, the calculated radius of the "atom" with such "heavy electrons" would be much smaller than the ordinary atom. This also proves that the assumptions of such a calculation fail: the heavy fermions (in condensed matter physics) are the result of the collective action of many atoms on the electron and its mass.
So the large mass of the heavy fermions is only appropriate for questions about physics at long distances - much longer than the ordinary atom. If you look at very short distances - a would-be tiny atom with the heavy fermion - you cannot use the long-distance or low-energy effective approximations of condensed matter physics. You have to return to the more fundamental, short-distance or high-energy description which sees electrons again.
At any rate, you will find out that there can be no supertiny atoms created out of the effective particles such as heavy fermions. The validity of all such phenomenological effective theories - such as those with heavy fermions - is limited to phenomena at distances longer than a certain specific cutoff and highly sub-atomic distances surely violate this condition, so one must use a more accurate theory than this effective theory, and in those more effective theories, most of the fancy emergent condensed matter objects disappear.
Non-relativistic effective theories
Just a disclaimer for particle physicists: in this condensed matter setup, we are talking about non-relativistic theories so the maximum allowed energy $E$ of quasiparticles doesn't have to be $pc$ where $p$ is the maximum allowed momentum in the effective theory. In other words, we can't assume $v/c=O(1)$. Quite on the contrary, the validity of such effective theories in condensed matter physics typically depends on the velocities' being much smaller than the speed of light, too.
So the mass of the heavy fermions is much greater than $m_0$ which would make $m_e c^2$ much greater than $m_0 c^2$; however, the latter is not a relevant formula for energy in non-relativistic theories. Instead, $p^2/2m_e$, which is (for heavy fermions) much smaller than the kinetic energy of electrons, is relevant. The maximum allowed $p$ of these quasiparticles is much larger than $\hbar/r_{\rm Bohr}$ - the de Broglie wavelength must be longer than the Bohr radius. That makes $p^2/2m_e$ really tiny relatively to the Hydrogen ionization energy.