Under the assumption that arriving customers are turned away when there are three customers in the system, this is a $M/M/1/3$ queue, that is, a single-server queueing system with arrival process Poisson, i.i.d. exponentially distributed service times, and a finite buffer size of $3$. Let $X(t)$ be the number of customers in the system at time $t$. Then $X=\{X(t):t\in\mathbb R_+\}$ is a continuous-time Markov chain with state space $S = \{0,1,2,3\}$ and transition rates
$$
q_{ij} =
\begin{cases}
2,& j=i+1\\
10,& j=i-1\\
0,& \text{otherwise}.
\end{cases}
$$
Let $P(t)$ be the matrix with entries $$[P(t)_{ij}] = \mathbb P(X(t) = j\mid X(0) = i).$$
Then $P$ satisfies the forward Kolmogorov equation $$P'(t) = P(t)Q(t), $$ with solution given by the matrix exponential $$P(t) = e^{tQ}.$$ (This may be verified by the Chapman-Kolmogorov equation $P(s+t) = P(s)P(t)$.) Since each row of $P(t)$ sums to $1$ (as it is a stochastic matrix), it follows that each row of $Q$ sums to $0$. The nondiagonal entries of $Q$ are precisely the transition rates $q_{ij}$, so the diagonal entries satisfy $$q_{ii} = -\sum_{j\in S\setminus\{i\}}q_{ij}. $$
Hence,
$$
Q = \begin{pmatrix}
-2 & 2 & 0 & 0\\
10 & -12 & 2 & 0\\
0 & 10 & -12 & 2\\
0 & 0 & 10 & -10.
\end{pmatrix}
$$
If $\pi$ is a probability distribution on $S$, let $\{X_\pi(t):t\in\mathbb R_+\}$ denote the process $X$ conditioned on $X(0)\sim\pi$. If $X_\pi(t)$ is a stationary process, that is, $\{X_\pi(s+t):t\in\mathbb R_+\}\stackrel d=\{X_\pi(t):t\in\mathbb R_+\}$ for all $s\geqslant 0$, then $\pi$ is a stationary distribution. From the balance equations $$\sum_{j\in S\setminus\{ i\}} \pi_{i}q_{ij} = \sum_{j\in S\setminus\{ i\}} \pi_{j}q_{ji},$$ it can be verified that $\pi$ is a stationary distribution iff $\pi Q=0$. In this example, $\pi Q=0$ is the system of linear equations
\begin{align}
-2\pi_0 + 10\pi_1 &= 0\\
2\pi_0 -12\pi_1 + 10\pi_2 &= 0\\
2\pi_1 -12\pi_2 + 10\pi_3 &= 0\\
2\pi_2 -10\pi_3 &= 0
\end{align}.
These equations, along with $\sum_{i\in S}\pi_i = 1$ (since $\pi$ is a probability distribution), yield
$$\pi = \begin{pmatrix}\dfrac{125}{156}& \dfrac{25}{156}& \dfrac{5}{156}& \dfrac1{156} \end{pmatrix}. $$
The fraction of time spent in each state satisfies
$$\lim_{T\to\infty}\frac1T \int_0^T \mathsf 1_{\{X(t)=i\}}\,\mathsf dt = \pi(i).$$
This serves as a good approximation for the fraction of time spent in a state for large $t$, but computing the exact value of $\int_0^T \mathsf 1_{\{X(t)=i\}}\,\mathsf dt$ for a given $T$ is non-trivial and (especially for larger problems) is generally done by simulation.
