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For the language L = $\{\Sigma^*. 0 .\Sigma^5 . 1. \Sigma^*\}$

The NFA must have 8 states. Also, what would be the upper bound on the number of states of a DFA recognizing L.

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The NFA looks good, though you should have said that $\Sigma=\{0,1\}$. There is no upper bound on the number of states of a DFA recognizing $L$; did you mean lower bound? – Brian M. Scott Nov 8 '12 at 7:50
The minimal automaton of your language has 65 states. Thus any DFA accepting your language will have at least 65 states. – J.-E. Pin Feb 3 at 12:19

Delete the $\lambda$-edge from $q_1$ to $q_8$. Presently the automaton accepts every string over $\{0,1\}$.

In your question delete the set-brackets in $\{ \Sigma^* \cdots\Sigma^* \}$. In itself, $\Sigma^*$ is a set (of strings), so it is not necessary (in fact wrong) to add another `level' of sets.

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