Initially, we are dealing with a non-homogeneous Dirichlet problem:
\begin{equation} \begin{cases} q_t(z,t)-D(z,t)q_{zz}(z,t) = 0 & \text{for $0<z<L,t>0$} \\ q(0,t)=a & \text{for $0\leq t$} \\ q(L,t)= 28\cdot 10^{-6} [kg/kg]& \text{for $0\leq t$} \\ q(z,0)= 28\cdot 10^{-6} [kg/kg]& \text{for $0\leq z\leq L$}\\ \end{cases} \end{equation}
We perform a variable transformation to homogenize our problem and compute its solution.
\begin{equation} \begin{cases} u_t(z,t)-D(z,t)u_{zz}(z,t) = 0 & \text{for $0<z<L,t>0$} \\ u(0,t)= 0 & \text{for $0\leq t$} \\ u(L,t)= 0 & \text{for $0\leq t$} \\ u(z,0)= 28\cdot 10^{-6}(1-\frac{z}{L})+a(\frac{z}{L}-1) & \text{for $0\leq z\leq L$}\\ \end{cases} \end{equation}
Considering $u(z,t)=M(z)N(t)$, we have $u_{zz}=M''(z)N(t)$ and $u_t=M(z)N'(t)$. Hence, we can express and solve individually for $M''(z)-\lambda M(z)=0$ and $N'(t)=D\lambda N(t)$. If $\lambda = 0$ or $\lambda > 0$, the initial conditions would not be satisfied, so they cannot be solutions. If $\lambda=-\omega ^2 <0$, we find the individual solutions for $M(z)$ and $N(t)$ as $M_n(z)=A_n \sin(\frac{n \pi}{L}z)$ and $N_n(t)=B_n e^{-D\frac{n^2\pi^2}{L}t}$, respectively. Therefore:
$$u(z,0)= \sum ^\infty _{n=1} C_n \sin(\frac{n \pi z}{L}) = 28 \cdot 10^{-6} (1-\frac{z}{L})+a(\frac{z}{L}-1)$$
And thus, the solution for $u(z,t)$ is expressed as:
\begin{equation} u(z,t)= \sum ^\infty _{n=1} \left[ \frac{2}{L} \int_{0}^{L} dz \left( 28 \cdot 10^{-6} (1-\frac{z}{L})+a(\frac{z}{L}-1)\right) \right] \sin(\frac{n \pi z}{L}) e^{\frac{Dn^2\pi^2}{L^2}t} \end{equation}
Now, I am uncertain about the procedure for undoing the variable transformation employed to homogenize the problem and obtain the final solution for q(z,t).
\begin{equation} q(z,t)= ?? + \sum ^\infty _{n=1} \left[ \frac{2}{L} \int_{0}^{L} dz \left( 28 \cdot 10^{-6} (1-\frac{z}{L})+a(\frac{z}{L}-1)\right) \right] \sin(\frac{n \pi z}{L}) e^{\frac{Dn^2\pi^2}{L^2}t} \end{equation}