The usual approach is to is to give the DFA a state for each set of states in the NDFA. In your case those are $t_\varnothing,t_0,t_1,t_2,t_{01},t_{02},t_{12}$, and $t_{012}$. That is, $t_{02}$, for instance, is a state of the DFA corresponding to the set $\{s_0,s_2\}$ of states of the NDFA.
The NDFA has a unique initial state, $s_0$, so its set of initial states is $\{s_0\}$, and we make $t_0$ the initial state of the DFA. An input of $0$ in state $s_0$ can take you to either $s_1$ or $s_2$ in the NDFA, so $\{s_1,s_2\}$ is the set of possible resulting states; this corresponds to $t_{12}$ in the DFA that we’re building, so we give it the transition $t_0\;\stackrel{0}\longrightarrow\; t_{12}\;.$
An input of $1$ to the NDFA in state $s_0$ takes it only to $s_0$, so $\{s_0\}$ is the set of possible resulting states, corresponding to $t_0$, and we must have a transition $t_0\;\stackrel{1}\longrightarrow\;t_0$ in the DFA.
What happens if I’m in one of the states in the set $\{s_1,s_2\}$ in the NDFA and get an input of $0$? If I’m in $s_1$, I go to $s_1$, and if I’m in $s_2$, no further computation is possible. The only $0$ transition from either of these states goes to $s_1$, so the set of possible resulting states is just $\{s_1\}$, and in the DFA the $0$ transition from $t_{12}$ must be to $t_1$: $t_{12}\;\stackrel{0}\longrightarrow\;t_1\;.$ Similar reasoning gives you the DFA transition $t_{12}\;\stackrel{1}\longrightarrow\;t_2\;.$
Now we need to compute the DFA $0$ and $1$ transitions for the states $t_1$ and $t_2$. You should have no trouble with $t_1$, but $t_2$ introduces a new wrinkle: neither input goes anywhere. This just means that the set of possible resulting states is $\varnothing$, the empty set, and the DFA transitions are $t_2\;\stackrel{0}\longrightarrow\;t_\varnothing$ and $t_2\;\stackrel{1}\longrightarrow\;t_\varnothing\;.$
Now that $t_\varnothing$ has appeared, we need to calculate the $0$ and $1$ transitions from it; they should be pretty clear after what we just did for $t_2$.
When you’ve filled in the blanks, you’ll find that the states $t_{01},t_{02}$, and $t_{012}$ never arose; you can still fill in transitions for them, but you need not, since these states are inaccessible from the initial state $t_0$.
One last task remains: to determine the acceptor state(s) of the DFA. That’s easy: if a set of NDFA states includes at least one acceptor state, the corresponding DFA state is an acceptor state. Here that means that $t_1,t_{12},t_{01}$, and $t_{012}$ are acceptor states, though we can ignore the last two, since they’re inaccessible.