= Solution
Retain the inferred mobilities and assume one ionized donor per added <arsenic> atom, negligible compensation, the same <band gap>, and the nondegenerate <semiconductor mass-action law>. Relative to the intrinsic sample, a conductivity factor $f$ requires
$$
n_e+n_h=2fn_i,\qquad n_en_h=n_i^2.
$$
For complete donor ionization, <charge neutrality> gives $N_D=n_e-n_h$, so
$$
\boxed{N_D=2n_i\sqrt{f^2-1}\simeq2fn_i
\simeq7.64\times10^{23}\ {\rm m^{-3}}
=7.64\times10^{17}\ {\rm cm^{-3}}\quad(f=1000).}
$$
The host atom density, using $M_{\rm Ge}\simeq72.63\ {\rm g\,mol^{-1}}$, is
$$
n_{\rm Ge}=\frac{5.32\times10^6\ {\rm g\,m^{-3}}}{72.63\ {\rm g\,mol^{-1}}}N_{\rm A}
\simeq4.41\times10^{28}\ {\rm m^{-3}}.
$$
Thus the required atomic fraction is
$$
\boxed{\frac{N_D}{n_{\rm Ge}}\simeq1.73\times10^{-5}
\simeq17.3\ \text{atomic ppm},}
$$
approximately one donor per $5.8\times10^4$ host atoms. The <electron> density must rise to about $2000n_i$, not $1000n_i$, because the original intrinsic sample conducted through both <electrons> and holes. This is an estimate under unchanged mobility: impurity scattering, incomplete ionization, compensation and band-edge changes can alter it. Here $n_e/N_c\simeq0.12$, so the nondegenerate treatment is an approximation rather than an arbitrarily dilute limit.
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