= Carrier density and the conductivity-ratio assumption
The atomic <number density> is
$$
n_{\rm at}=\frac{5300\,\mathrm{kg\,m^{-3}}}{0.073\,\mathrm{kg\,mol^{-1}}}N_A=4.372\times10^{28}\,\mathrm{m^{-3}}.
$$
One fully ionised single-electron donor per million atoms gives
$$
\boxed{N_D=4.372\times10^{22}\,\mathrm{m^{-3}}},\qquad
\boxed{N_D/n_i=1.46\times10^3},\quad n_i=3\times10^{19}\,\mathrm{m^{-3}}.
$$
The usual simplified answer treats conductivity as proportional to <Electron> concentration and assumes unchanged <electron mobility>, giving \b[about $1.46\times10^3$ times the undoped conductivity]. There is a qualification: the <conductivity ratio of doped and intrinsic semiconductors> is not fixed by <Electron> counts alone. Intrinsic material also has $p=n_i$, whereas the heavily n-type sample has $n\simeq N_D$ and $p\simeq n_i^2/N_D$. Therefore
$$
\boxed{\frac{\sigma_d}{\sigma_i}\simeq
\frac{N_D\mu_{e,d}}{n_i(\mu_{e,i}+\mu_{h,i})}}.
$$
With unchanged <Electron> mobility and negligible intrinsic hole conductivity this reduces to $1.46\times10^3$; with equal, unchanged <Electron> and hole mobilities it is about $7.29\times10^2$. The problem supplies no mobilities, so the density-ratio estimate requires its stated approximation. Ionised impurities may additionally reduce mobility.
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