= Solution
The sign must become positive, so use p-type <acceptor impurities>, such as boron substituting for tetravalent silicon. Assume uncompensated, fully ionized acceptors of density $N_A$, nondegenerate carriers and equilibrium throughout the interval. <Charge neutrality> and <mass> action give
$$
p-n=N_A,\qquad np=n_i^2,\qquad
\boxed{p=\frac{N_A+\sqrt{N_A^2+4n_i^2}}2,\quad n=\frac{n_i^2}{p}}.
$$
For a nearly constant hole-dominated <Hall coefficient>, the dopant plateau must persist up to the highest <temperature>. Its necessary concentration scale is therefore
$$
\boxed{N_A\gg n_i(500)\simeq4.0\times10^{20}\ \mathrm{m^{-3}}
=4.0\times10^{14}\ \mathrm{cm^{-3}}}.
$$
The order-of-magnitude onset is $N_A\sim4\times10^{20}\ \mathrm{m^{-3}}$, but equality is not a flat plateau: at $N_A=n_i(500)$, $p/N_A=(1+\sqrt5)/2=1.618$, changing the hole-only <Hall coefficient> by about 38 percent. The word approximately does not define a unique sharp minimum without a tolerance.
For $N_A\gg n_i$, $p\simeq N_A[1+(n_i/N_A)^2]$, so the hole-only $R_H\simeq1/(eN_A)$ varies by a fraction of order $(n_i/N_A)^2$. An illustrative one-percent tolerance in that approximation is obtained with $N_A\simeq10n_i(500)=4.0\times10^{21}\ \mathrm{m^{-3}}$. This gives $R_H\simeq+1.56\times10^{-3}\ \mathrm{m^3\,C^{-1}}$ and less than one percent hole-density-induced variation. It is an explicit plateau criterion, not an exact minimum deduced from the unspecified word approximately.
At 300 K the minority <electron> concentration is
$$
\boxed{n(300)\simeq\frac{[9.99\times10^{15}]^2}{N_A}
=2.50\times10^{11}\left(\frac{4.0\times10^{20}\ \mathrm{m^{-3}}}{N_A}\right)\ \mathrm{m^{-3}}}.
$$
Thus the onset-scale estimate gives about $2.5\times10^{11}\ \mathrm{m^{-3}}$, whereas the explicit ten-times-larger plateau choice gives $2.5\times10^{10}\ \mathrm{m^{-3}}$.
The <acceptor density for a stable Hall sign> can be more restrictive if both carriers matter. With mobility ratio $r_\mu=\mu_e/\mu_h>1$, positive Hall sign requires $p>r_\mu n_i$, or $N_A>n_i(r_\mu-r_\mu^{-1})$. In the strongly extrinsic limit the full two-carrier Hall correction scales as $(1+r_\mu)^2(n_i/N_A)^2$, not just $(n_i/N_A)^2$. Hence a true measured-Hall tolerance can require a larger dopant density than the hole-only illustration. Mobilities and their <temperature> dependence are not supplied, so they cannot be used to manufacture a unique sharper minimum.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2015-intrinsic-carriers-acceptor-plateau.png]
{title=Intrinsic carrier activation and the hole-only Hall plateau for two explicitly chosen acceptor densities; concentrations are model estimates, not extra measurements}
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