Solution:

Call a real number very large if $x \in [1000, 1008]$, very small if $x \in \left[0, \frac{1}{1000}\right]$, and medium-sized if $x \in \left[\frac{1}{8}, 8\right]$. Every number Rachel is ever holding after at most $8$ steps will fall under one of these categories. Therefore the main contribution to $E$ will come from the probability that Rachel is holding a number at least $1000$ at the end.

Note that if her number ever becomes medium-sized, it will never become very large or very small again. Therefore the only way her number ends up above $1000$ is if the sequence of moves consists of $x \rightarrow x+1$ moves and consecutive pairs of $x \rightarrow x^{-1}$ moves. Out of the $256$ possible move sequences, the number of ways for the number to stay above $1000$ is the number of ways of partitioning $8$ into an ordered sum of $1$ and $2$, or the ninth Fibonacci number $F_{9} = 34$.

Therefore
$$\frac{34}{256} \cdot 1000 \leq E \leq \frac{34}{256} \cdot 1000 + 8$$
where $\frac{34}{256} \cdot 1000 \approx 132.8$. Furthermore, the extra contribution will certainly not exceed $7$, so we get that $\left\lfloor\frac{E}{10}\right\rfloor = 13$.