= Solution
{c}
The plateau in both source plots is approximately $10^{18.5}\,\mathrm{m^{-3}}$, so
$$
\boxed{N_D\simeq3\times10^{18}\,\mathrm{m^{-3}}.}
$$
These are graphical estimates, not high-precision fitted material parameters. To extract activation <energies> consistently with part (a), retain the <temperature> powers in the <effective density of states> factors.
On the <semiconductor carrier freeze-out> tail, put $x=1/T$. Then
$$
\log_{10}n=\text{constant}-\frac34\log_{10}x
-\frac{E_i}{2k_B\ln10}x.
$$
For example, the right-hand graph gives roughly $\log_{10}[n/(\mathrm{m^{-3}})]\simeq17.7$ at $x=0.10\,\mathrm{K^{-1}}$ and $13.0$ at $x=0.30\,\mathrm{K^{-1}}$. Subtract the known prefactor contribution before interpreting the slope:
$$
E_i\simeq2k_B\ln10\,
\frac{4.7-(3/4)\log_{10}3}{0.20\,\mathrm{K^{-1}}}
\simeq8.6\times10^{-3}\,\mathrm{eV}.
$$
Thus $\boxed{E_i\simeq9\,\mathrm{meV}}$, to the accuracy of the drawing. Ignoring the weak prefactor gives about $9.3\,\mathrm{meV}$, a similar rough estimate.
For the intrinsic branch, use
$$
\ln\frac{n(T_2)}{n(T_1)}=\frac32\ln\frac{T_2}{T_1}
+\frac{E_g}{2k_B}\left(\frac1{T_1}-\frac1{T_2}\right).
$$
The left-hand graph gives $n(100\,\mathrm K)\simeq2.5\times10^{21}\,\mathrm{m^{-3}}$ and $n(500\,\mathrm K)\simeq1.5\times10^{24}\,\mathrm{m^{-3}}$, both far above $N_D$. Therefore
$$
E_g\simeq\frac{2k_B\left[\ln600-(3/2)\ln5\right]}
{(1/100-1/500)\,\mathrm{K^{-1}}}
\simeq0.086\,\mathrm{eV},\qquad
\boxed{E_g\simeq0.09\,\mathrm{eV}.}
$$
Here $k_B=8.6173\times10^{-5}\,\mathrm{eV\,K^{-1}}$. Omitting the $T^{3/2}$ prefactor over this wide <temperature> interval would instead give approximately $0.14\,\mathrm{eV}$, so it is not a negligible correction for the <band-gap extraction from intrinsic carrier activation>. The source graph warrants an approximate narrow <band gap>, rather than several significant figures.
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