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Solution

The automaton has two “phases” connected by the path recognizing a→b→c. Each time the substring “abc” is seen, the DFA moves to the other phase, toggling parity. The start phase (no “abc” yet) is accepting (count 0), and after every additional “abc” the parity flips; therefore the accepting states correspond to an even number of occurrences of “abc”. Options 1 and 2 describe Σ* (all strings), and option 3 does not match the transitions.