Let $C$ be an arbitrary set in $X$. By Caratheodory's theorem the $\mu^{*}$-measurable sets form a $\sigma$-algebra, and so $B$ is $\mu^*$-measurable.
Therefore $\mu^*(C) = \mu^*(C \cap B) + \mu^*(C \cap B^{c}).$ Since $E \subseteq B$ we have that $\mu^*(C \cap E) \leq \mu^*(C \cap B)$.
And since $\mu^*(B \setminus E) = 0$ we have $\mu^*(C \cap (B \setminus E)) = 0$.
Finally, we use the fact that $E^{c} = B^{c} \cup (B \setminus E)$, which tells us that $\mu^*(C \cap E^{c}) \leq \mu^*(C \cap B^{c}) + \mu^*(C \cap (B \setminus E)) = \mu^*(C \cap B^{c})$.
Putting this all together
\begin{equation}
\mu^*(C) = \mu^*(C \cap B) + \mu^*(C \cap B^{c}) \geq \mu^*(C \cap E) + \mu^*(C \cap E^{c}),
\end{equation}
implying that $E$ is $\mu^*$-measurable.
We begin with a small lemma that in fact highlights part of what has already been proved in Exercise 18 item b.).
Lemma: For any $E\subset X$ there exists $B\in \mathcal{A}_{\sigma\delta}$ such that $E\subset B$ and $\mu^*(B) = \mu^*(E)$.
Proof
From Exercise 18 item a.) we know that
For any $E\subset X$ and $\epsilon > 0$ there exists $A\in \mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(E)\leq \mu^*(A) \leq \mu^*(E) + \epsilon$.
So, for each $n\in\mathbb{N}$, $n>0$, let $A_n\in \mathcal{A}_\sigma$ with $E\subset A_n$ and $\mu^*(E)\leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$.
Then let $B=\bigcap_{n=1}^\infty A_n$. Then we have $B\in \mathcal{A}_{\sigma\delta}$ and $E\subset B$. Moreover, for all $n\in\mathbb{N}$, $n>0$, $B\subset A_n$ and
$$\mu^*(E)\leq \mu^*(B) \leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$$
So, $\mu^*(B)=\mu^*(E)$.
Remark: $E$ don't need to be measurable. And as a consequence of item Exercise 18 item b.) $\mu^*(B\setminus E)$ may not be zero.
Exercise 19 - Let $\mu^*$ be an outer measure on $X$ induced from a finite premeasure $\mu_0$. If $E\subset X$, define the inner measure of $E$ to be $\mu_*(E) = \mu_0(X) - \mu^*(E^c)$. Then $E$ is $\mu^*$-measurable iff $\mu^*(E) = \mu_*(E)$ (Use Exercise 18).
First note that since $\mu_0$ is a finite premeasure, we have that for all $A\subset X$, $\mu^*(A)<+\infty$.
($\Rightarrow$) Suppose $E\subset X$. Note that $E^c=X\setminus E \subset X$. So,since $E$ is $\mu^*$-measurable, we have
$$\mu^*(X) = \mu^*(X\cap E) + \mu^*(X\cap E^c)=\mu^*(E) + \mu^*(E^c)$$ then, since $\mu^*(E^c)<+\infty$ and $\mu^*(X)=\mu_0(X)$, we have
$$\mu^*(E) = \mu^*(X) - \mu^*(E^c)=\mu_0(X)- \mu^*(E^c)=\mu_*(E)$$
($\Leftarrow$) If $\mu^*(E) =\mu_*(E)$ then we have, since $\mu^*(X)=\mu_0(X)$,
$$\mu^*(E) =\mu_*(E)=\mu_0(X)- \mu^*(E^c) = \mu^*(X) - \mu^*(E^c)$$
So we can conclude that
$$\mu^*(X) = \mu^*(E)+ \mu^*(E^c) \tag{1}$$
Now we apply our lemma to $E$ and $E^c$. Let $B, D \in \mathcal{A}_{\sigma\delta}$ such that $E\subset B$ and $\mu^*(B) = \mu^*(E)$ and $E^c\subset D$ and $\mu^*(D) = \mu^*(E^c)$. From $(1)$, we have
$$\mu^*(X) = \mu^*(B)+ \mu^*(D) \tag{2}$$
On the other hand, since $D \in \mathcal{A}_{\sigma\delta}$, we have that $D$ is $\mu^*$-measurable, so
$$ \mu^*(X) = \mu^*(D)+ \mu^*(D^c) \tag{3}$$
From $(2)$ and $(3)$, we get
$$ \mu^*(D)+ \mu^*(D^c)= \mu^*(B)+ \mu^*(D) $$
Since $\mu^*(D) <\infty$, we have
$$\mu^*(D^c) = \mu^*(B) \tag{4} $$
Note that since $E^c\subset D$ we have that $D^c\subset E$. So we actually have
$$D^c\subset E \subset B \tag{5}$$
and, since $D$ are $\mu^*$-measurable, $D^c$ is also $\mu^*$-measurable.
So
$$\mu^*(B)=\mu^*(D^c) + \mu^*(B\setminus D^c) \tag {6}$$
Since $\mu^*(D^c)<+\infty$ (and $\mu^*(B)<+\infty$), we have from $(4)$ and $(6)$ that
$$\mu^*(B\setminus D^c)=0$$
But from $(5)$ we have $B \setminus E\subset B\setminus D^c$, so $$\mu^*(B\setminus E)=0$$ By exercise 18 item b.), $E$ is $\mu^*$-measurable.
Best Answer
The definition of $\mu^*$ is as follows:$\mu^*(E)=\inf\{\sum_{j=1}^\infty \mu_0(E_j)|A\subseteq \cup_{j=1}^\infty E_j,E_j\in A\}$.
Therefore for any $E$, $\mu^*(E)+\epsilon > \mu^*(E)$. Hence, $\mu^*(E)+\epsilon$ is larger than the infimum. Therefore by the definition of infimum there exists $\{E_j\}_{j\in N}, E_j\in A$ such that $A\subseteq \cup_{j=1}^\infty E_j$ and $\mu^*(E)+\epsilon >\sum_{j=1}^\infty \mu_0(E_j) $.
Note: $\cup_{j=1}^\infty E_j\in A_\sigma$ and $\mu^*(\cup_{j=1}^\infty E_j)\leq \sum_{j=1}^\infty \mu_0(E_j)$.
Hence, $\mu^*(\cup_{j=1}^\infty E_j)<\mu^*(E)+\epsilon$.