DFA drawing for binary string with substrings of minimum length 3 with at least two zeroes in each substring
The condition that all words have length $\geqslant 3$ makes the computation of the minimal DFA much more difficult. Thus let us first consider the language 
$$
K = \{ u \in A^* \mid \text{every substring of length 3 of $u$ contains at least two $0$} \}
$$
where $A = \{0,1\}$. The forbidden patterns of the words of $K$ are $11$ and $101$, so that $K$ is the complement of the language $A^*(11 + 101)A^*$. But there is a nicer way to write $K$, using the comment of Luke Mathieson. Indeed, apart from the last two letters, every $1$ has to be followed by two $0$. This leads to the regular expression
$K = (0 + 100)^*(\varepsilon + 1 + 10)$. I let you compute the minimal DFA of $K$ which has only three states (plus a sink state if you want a complete DFA).

Now $L = A^3A^* \cap K$ and you can use standard algorithms to get its minimal DFA
(9 states, 10 if you count the sink state).