A Markov Chain {$X_n,n\geq0$} with states $0,1,2$, has the transition probability matrix $$\begin{bmatrix}\frac{1}{2}&\frac{1}{3}&\frac{1}{6}\\0&\frac{1}{3}&\frac{2}{3}\\\frac{1}{2}&0&\frac{1}{2}\end{bmatrix}$$ If $P(X_0=0)=P(X_0=1)=\frac{1}{4}$, find $E[X_3]$.
I'm not so sure, because $X_3$ it is outside the states $$P(X_{n+1}=j|X_n=i)$$ $$E[X_3]=0P(X_3=0)+1P(X_3=1)+2P(X_3=2)$$ that's what they want to mean?
Yes, this is the correct way to calculate $E[X_3]$ $$E[X_3]=0P(X_3=0)+1P(X_3=1)+2P(X_3=2)$$ The $3$ corresponds to the temporal dimension, not the spatial dimension, which can be any $n$ from $0$ onward.
You have sufficient information to calculate the probabilities of being in each spatial state at time $3$. You are given the initial probabilities $P(X_0 = i_0)$ and the transition probabilities $P(X_{n+1} = i_{n+1} | X_n = i_n)$.