Substituting the $u=e^{\lambda y}$ into the equation, we have
\[e^{\lambda y} \left( \lambda^2 + 2i \alpha \lambda + \beta - \alpha^2 \right) = 0.\]
So, let's solve the quadratic equation:
\[\mathcal{D} = -4 \alpha^2 - 4 \beta + 4 \alpha^2 = -4 \beta, \quad \lambda_{1,2} = \frac{-2i \alpha \pm 2 \sqrt{-\beta}}{2}\]
There are required relations for the constants:
\[
\begin{cases}
A_1 e^{-\lambda_{11}b/a} + B_1 e^{-\lambda_{12}b/a} = A_2 e^{\lambda_{21}(1-b/a)}
+ B_2 e^{\lambda_{22}(1-b/a)}\\
A_1 + B_1 = A_2 + B_2 \\
\lambda_{11} A_1 + \lambda_2 B_1 = \lambda_{21} A_2 + \lambda_{22} B_2\end{cases}\]
The 4th equation could be simplified to
\[ u'_I(-b/a) = u_{II}'(1-b/a) \quad \Rightarrow \quad A_1 \lambda_{11} e^{-\lambda_{11}b/a} + B_1 \lambda_{12} e^{-\lambda_{12}b/a} = A_2 \lambda_{21} e^{\lambda_{21}(1-b/a)} + B_2 \lambda_{22} e^{\lambda_{22}(1-b/a)}\]
To find the $E_F$ we have to consider the "box" with $N$ occupied levels. As $k= \pi n a /L$ and we have two different spin directions, $k_\text{max}=\pi N a / 2L$. Thus, the energy of the highest occupied level
\[E_F = \frac{\hbar^2 k_\text{max}^2}{2ma^2} = \frac{\pi^2 \hbar^2 N^2}{8 m L^2}\]
In the program we work with ondimensionalized energy $\varepsilon$. To find the energy $E$ that corresponds to it we have to use such an equation
\[ E = \frac{\varepsilon b}{a} U_0 = \varepsilon \frac{k_0^2 \hbar^2}{2ma^2}\]
\[
\begin{split}
p &= \sum_{E \leq E_v} \left[ g(\Delta E, k) - F(g(\Delta E,k),E) \right] = \sum_{E \leq E_v} \left[ g_v (\Delta E,k) - \frac{g_v (\Delta E,k)}{e^{\frac{E-E_F}{k_\text{B}T}}+1} \right] \simeq \sum_{E \leq E_v} g_v e^{\frac{E-E_F}{k_\text{B}T}} \simeq\\&\simeq \int\limits_{-\infty}^{E_v}\frac{L}{\pi a \alpha k} e^{\frac{E - E_F}{k_\text{B}T}} dE = \frac{2L}{\pi a} e^{\frac{E_v- E_F}{k_\text{B}T}} \int\limits_0^{\infty} e^{-\frac{\beta k^2}{k_BT}}dk = \frac{L}{a} e^{\frac{E_v - E_F}{k_\text{B}T}} \sqrt{\frac{k_B T}{\pi \beta}}.
\end{split}\]
In the pure silicon $\rm Si$ we have $p=n$, so in the doped silicon there is $p = qn$. Changing atoms only shift the Fermi level, so $\sqrt{pn}=n_i$ and
\[ p =n_i \sqrt{q}, \quad n = \frac{n_i}{\sqrt{q}}.\]
Now, let's satisfy the electronutrality condition:
\[ p - \frac{cL}{a} = n\]
and get the answer:
\[ c = \left( \sqrt{q} + \frac{1}{\sqrt{q}} \right) \cdot 2.6 \cdot 10^{-11} \simeq 2.6 \cdot 10^{-8}. \]