Let $\delta$ denote the transition function and $\hat{\delta}$ denote the extended transition function of the $\epsilon$-NFA whose transition table is given below: Then $\hat{\delta}(q_2, aba)$ is

To compute the extended transition function $\hat{\delta}(q_2, aba)$ for the given $\epsilon$-NFA, we process the string symbol-by-symbol using $\epsilon$-closures:

1. Initial state closure:

   $$\epsilon\text{-closure}(q_2) = \{q_0, q_2\}$$

   (Since $\delta(q_2, \epsilon) = \{q_0\}$ and $\delta(q_0, \epsilon) = \{q_2\}$).

2. On reading input symbol 'a':

   $$\delta(\{q_0, q_2\}, a) = \delta(q_0, a) \cup \delta(q_2, a) = \{q_1\} \cup \emptyset = \{q_1\}$$

   Taking $\epsilon$-closure:

   $$\epsilon\text{-closure}(\{q_1\}) = \{q_0, q_1, q_2\}$$

   (Since $q_1 \xrightarrow{\epsilon} q_2 \xrightarrow{\epsilon} q_0 \xrightarrow{\epsilon} q_2$).

3. On reading input symbol 'b':

   $$\delta(\{q_0, q_1, q_2\}, b) = \delta(q_0, b) \cup \delta(q_1, b) \cup \delta(q_2, b) = \{q_0\} \cup \{q_3\} \cup \emptyset = \{q_0, q_3\}$$

   Taking $\epsilon$-closure:

   $$\epsilon\text{-closure}(\{q_0, q_3\}) = \epsilon\text{-closure}(q_0) \cup \epsilon\text{-closure}(q_3) = \{q_0, q_2\} \cup \{q_3\} = \{q_0, q_2, q_3\}$$

4. On reading input symbol 'a':

   $$\delta(\{q_0, q_2, q_3\}, a) = \delta(q_0, a) \cup \delta(q_2, a) \cup \delta(q_3, a) = \{q_1\} \cup \emptyset \cup \emptyset = \{q_1\}$$

   Taking $\epsilon$-closure:

   $$\epsilon\text{-closure}(\{q_1\}) = \{q_0, q_1, q_2\}$$

Thus, $\hat{\delta}(q_2, aba) = \{q_0, q_1, q_2\}$, which corresponds to option (C).