When the input is constant during one time period is also known as zero-order hold discretization and can be calculated with
$$
\begin{bmatrix}
A_d & B_d \\ 0 & 1
\end{bmatrix} =
\exp\left(
\begin{bmatrix}
A & B \\ 0 & 0
\end{bmatrix} h
\right),
$$
and $C_d = C$. The matrix exponential can be calculated with
$$
\exp M = \sum_{n=0}^\infty \frac1{n!} M^n.
$$
In am not sure if the question is also considering similarity transformations, otherwise every option where $C \neq C_d$ can already be omitted. When assuming that similarity transformations are not used then one only has to check two options, namely C) and E).
Calculating the discretization for C) gives
$$
\begin{bmatrix}
e^3 & 3\,e^3 & e^3 - 1 \\
0 & e^3 & 0 \\
0 & 0 & 1
\end{bmatrix} =
\exp\left(
\begin{bmatrix}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix} 3
\right),
$$
and calculating the discretization for E) gives
$$
\begin{bmatrix}
e^3 & 3\,e^3 & e^6 - 2\,e^3 + 1 \\
0 & e^3 & 0 \\
0 & 0 & 1
\end{bmatrix} =
\exp\left(
\begin{bmatrix}
1 & 1 & e^3 - 1 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix} 3
\right),
$$
from this it can be concluded that C) is the answer.
What you basically have to determine is which modes are controllable and observable. Checking the rank of the controllability and observability matrices won't tell you which modes are controllable and observable, only whether the entire system is minimal (and to some extend how many modes are not present). A better method in this case would be the Hautus lemma. However a direct application of this would this require you to check the rank of eight matrices (all four eigenvalues with $B$ for controllability and $C$ for observability). This can be reduced significantly by using the similarity transformation $\hat{x} = V^{-1}\,x$, which gives
\begin{align}
\hat{A} &= V^{-1}\,A\,V = \Lambda, \\
\hat{B} &= V^{-1}\,B, \\
\hat{C} &= C\,V.
\end{align}
Here $\hat{B}$ is twice the first column plus the last column of $V^{-1}$ and $\hat{C}$ is the first minus the second row of $V$, which gives
$$
\hat{B} = \begin{bmatrix}
1 & 0 & 1 & 1
\end{bmatrix}^\top, \quad
\hat{C} = \begin{bmatrix}
0 & 0 & 1 & 1
\end{bmatrix}.
$$
From this result it is relatively easy to see, without actually applying the Hautus test, which modes are controllable and observable since $\hat{A}$ is diagonal, so the system can be seen as four decoupled first order systems. The first component of the state $\hat{x}$ has the associated eigenvalue $-4$ and its dynamics does get an input passed through by $\hat{B}$ (its first component is nonzero), but does not appear in the output (the first component of $\hat{C}$ is zero), so this state component is controllable but not observable. Similarly it can be shown that the state with eigenvalue $-2$ not controllable and not observable, and the states with eigenvalues $0$ and $1$ are both controllable and observable.
So a minimal representation of the system would only contain the states which are both controllable and observable. This would be the states with eigenvalues $0$ and $1$, therefore answer B) would be the only option. However B) is not minimal, since by factoring out the 2 it is easy to see that it has also a zero at $1$, which would mean you could do pole zero cancelation. This would imply the system was not minimal, but earlier we showed that it is minimal. So all that remains is E) none of the above.
The minimal transfer function can be calculated relatively easily by considering only the controllable and observable parts. A minimal state space model would then become
$$
\left[\begin{array}{c|c}
A_m & B_m \\ \hline C_m
\end{array}\right] =
\left[\begin{array}{cc|c}
0 & 0 & 1 \\ 0 & 1 & 1 \\ \hline 1 & 1
\end{array}\right]
$$
Its corresponding transfer function can be calculated with $C_m\,(s\,I-A_m)^{-1}B_m$, where $s\,I-A_m$ is diagonal so its inverse is just taking the inverse of each diagonal element. Using this inverse gives the following expression for the transfer function
$$
H(s) = \frac{1}{s} + \frac{1}{s-1} = \frac{2\,s-1}{s(s-1)}.
$$
Best Answer
Your equations give: $$y = x_2\\sx_2=x_3\\sx_3=0$$ Which implies that: $$s^2y=0$$ Thus the transfer function is indeed zero.
Such systems are called "finite-memory" (particularly for discrete-time systems) and their matrices are nilpotent: $$\exists n | A^n = \mathbb{0} $$ Finite-memory systems have null output in a finite amount of time (as opposed to the usual, asymptotic behaviour of stable systems) when input is also null.