I transformed your NFA into a DFA as an exercise for my future test.
$\epsilon-closure\{S_0\}= \{S_0,S_1,S_3,S_4,S_5,S_6,S_7,S_9,S_{12}\}= A$
$1)$ From $A$ consuming $0$: $\{S_2, S_8 ,S_{13}\}$
$\epsilon-closure\{S_2, S_8 ,S_{13}\}= \{S_2,S_5,S_6,S_7,S_8,S_9,S_{11},S_{12},S_{13}\}=B$
From $A$ consuming $1$: $\{S_{10}\}$
$\epsilon-closure\{S_{10}\}= \{S_6,S_7,S_9,S_{10},S_{11},S_{12}\}=C$
$2)$ From $B$ consuming $0$: $\{S_2, S_8 ,S_{13}\}$
$\epsilon-closure\{S_2, S_8 ,S_{13}\}= B$
From $B$ consuming $1$: $\{S_{10}\}$
$\epsilon-closure\{S_{10}\}=C$
$3)$ From $C$ consuming $0$: $\{S_8 ,S_{13}\}$
$\epsilon-closure\{S_8 ,S_{13}\}= \{S_6,S_7,S_8,S_9,S_{11},S_{12},S_{13}\}=D$
From $C$ consuming $1$: $\{S_{10}\}$
$\epsilon-closure\{S_{10}\}= C$
$4)$ From $D$ consuming $0$: $\{S_8 ,S_{13}\}$
$\epsilon-closure\{S_8 ,S_{13}\}= D$
From $D$ consuming $1$: $\{S_{10}\}$
$\epsilon-closure\{S_{10}\}= C$
Then the initial state of the DFA is $A$ because it contain $S_0$.
The final states of the DFA are $B$ and $D$ because they contain $S_{13}$.
