Let me start with a little nitpicking:

- What on Earth do you mean by $\mu^*L-\epsilon E$ when you said that $L$ was a line bundle? You cannot add a line bundle and a divisor! So, let's assume that you said that $L$ was a Cartier divisor (or at least $\mathbb Q$-Cartier).
- It looks like you are assuming that Y is smooth, or at least that it is smooth along Z, but you didn't say so. If $Y$ is not necessarily smooth, then the exceptional set is not necessarily a divisor or irreducible, so your description of $E$ as
**the** exceptional divisor is totally unsubstantiated. So let's assume that $Y$ is $\mathbb Q$-factorial and that you want an exceptional effective $\mathbb Q$-divisor $E$ such that $\mu^*L-E$ is $\mathbb Q$-ample.

With that, you can do this:

Let $H$ be a very ample Cartier divisor on $X$ (Homework: why is there such a divisor?). Then $\mu_*H$ is a Weil divisor on $Y$ and if $Y$ is $\mathbb Q$-factorial, then it is $\mathbb Q$-Cartier and hence $\mu^*\mu_*H$ makes sense and is a $\mathbb Q$-Cartier divisor.

Then it is easy to see that $H-\mu^*\mu_*H$ is a $\mu$-very ample $\mathbb Q$-Cartier divisor and also that it is negative effective $\mu$-exceptional.
Finally, since $L$ is ample it follows that $a\mu^*L+H-\mu^*\mu_*H$ is ample for $a\gg 0$. (One could for example argue that $aL-\mu_*H$ is ample (actually nef is enough) for $a\gg0$ and hence $a\mu^*L+H-\mu^*\mu_*H$ is also ample).