One way is to take it in pieces. Clearly your automaton accepts $0^*$. What else gets it back to state $a$? Only $11^*00$, after which you can have any number of zeroes again. Thus, $0^*(11^*000^*)^*$ almost does the trick. However, like your automaton, it misses the possibility that an acceptable string can end in $1$ or $10$ if no $0$ immediately precedes the $1$. (For example, the words $11$ and $1110$ should be accepted.) Can you see how to adjust it (and your automaton) to reflect that possibility?
We can describe the set of valid strings as the strings built from the alphabet $V=\{a,b\}$ which contain at most one or two consecutive $a$'s.
These are strings
starting with zero or more $b$'s:$$b^*$$
followed by zero or more occurrences of $a$ or $aa$ each followed by one or more $b$'s: $$(ab^+|aab^+)^*$$
and terminating with zero, one or two $a$'s
$$(\varepsilon|a|aa)$$
We obtain
\begin{align*}b^*(ab^+|aab^+)^*(\varepsilon|a|aa)\tag{1}\end{align*}
$$ $$
Add-on: The regular expression (1) generates all valid words in a unique manner. In such cases we can use it to derive a generating function $$A(z)=\sum_{n=0}^\infty a_n z^n$$ with
$a_n$ giving the number of valid words of length $n$.
In order to do so all we need to know is the geometric series expansion since the $star$ operator
\begin{align*}
a^*=\left(\varepsilon|a|a^2|a^3|\cdots\right)\qquad\text{ translates to }\qquad 1+a+a^2+a^3+\cdots=\frac{1}{1-a}
\end{align*}
Accordingly $a^+=aa^*$ translates to $\frac{a}{1-a}$ and alternatives like $(\varepsilon|a|aa)$ can be written as $1+a+a^2$.
We obtain by translating the regular expression in the language of generating functions (and by mixing up somewhat the symbolic to provide some intermediate steps)
\begin{align*}
b^*\left(ab^+|aab^+\right)^*(\varepsilon|a|aa)
&\longrightarrow \quad \frac{1}{1-z}\left(\left.\frac{z^2}{1-z}\right|\frac{z^3}{1-z}\right)^*\left(1+z+z^2\right)\\
&\longrightarrow \quad \frac{1}{1-z}\left(\frac{z^2+z^3}{1-z}\right)^*(1+z+z^2)\\
&\longrightarrow \quad \frac{1}{1-z}\cdot\frac{1}{1-\frac{z^2+z^3}{1-z}}(1+z+z^2)\\
&\quad\quad=\frac{1+z+z^2}{1-z-z^2-z^3}
\end{align*}
We conclude: The number of valid words is given by the generating function $A(z)$
\begin{align*}
A(z)&=\frac{1+z+z^2}{1-z-z^2-z^3}\\
&=1+2z+4z^2+7z^3+13z^4+\color{blue}{24}z^5+44z^6+81z^7+\cdots
\end{align*}
The expansion was done with the help of Wolfram Alpha. We see that e.g. the number of valid words of length $5$ is $24$.
So, out of $2^5=32$ binary words of length $5$ there are $8$ invalid words marked blue in the table below.
\begin{array}{cccc}
\color{blue}{aaaaa}\qquad&ab\color{blue}{aaa}\qquad&ba\color{blue}{aaa}\qquad&bb\color{blue}{aaa}\\
\color{blue}{aaaa}b\qquad&abaab\qquad&b\color{blue}{aaa}b\qquad&bbaab\\
\color{blue}{aaa}ba\qquad&ababa\qquad&baaba\qquad&bbaba\\
\color{blue}{aaa}bb\qquad&ababb\qquad&baabb\qquad&bbabb\\
aabaa\qquad&abbaa\qquad&babaa\qquad&bbbaa\\
aabab\qquad&abbab\qquad&babab\qquad&bbbab\\
aabba\qquad&abbba\qquad&babba\qquad&bbbba\\
aabbb\qquad&abbbb\qquad&babbb\qquad&bbbbb\\
\end{array}
Best Answer
Zero is an even number. The string $aa$ contains an even number of $b$s. The empty string (which your RE matches) also contains an even number of $b$s.