It’s not quite right: the only word that puts the automaton in state $s_0$ is $\epsilon$, the empty word, and the only words that put the automaton in state $s_1$ are those of the form $01^*$. Thus, the desired regular expression is $01^*+\epsilon$ (or $01^*\cup\epsilon$, if you use that version of the notation). Any string that starts with $1$ ends up permanently in $s_2$, where it’s not accepted, and the same is true of any string with more than one $0$.
It’s not quite right: the only word that puts the automaton in state $s_0$ is $\epsilon$, the empty word, and the only words that put the automaton in state $s_1$ are those of the form $01^*$. Thus, the desired regular expression is $01^*+\epsilon$ (or $01^*\cup\epsilon$, if you use that version of the notation). Any string that starts with $1$ ends up permanently in $s_2$, where it’s not accepted, and the same is true of any string with more than one $0$.