The solution for $(c)$ is perfect up to $\mu^{*}(C_n\setminus E_n) = \mu^{*}(C_n) - \mu^{*}(E_n) \leq \epsilon/2^{n}$, so let's keep going from here.
It is easy to lose track of what we are doing with these kind of problems, so let's not forget that we are trying to approximate the set $E$ from outside, and that right now we only have an approximation for every $n$. The obvious thing to do is then to consider the union of the approximating sets: let $$C_{\epsilon} = \bigcup_{n = 1}^{\infty}C_n.$$
The $\epsilon$ in $C_{\epsilon}$ is there to stress the dependence upon $\epsilon$ of the $C_n$'s and hence of $C$. It is worth noticing that $C_{\epsilon} \in \mathcal{A}_{\sigma}$ being the countable union of countable unions of elements of $\mathcal{A}$.
For the next step to work you might want to state clearly in your proof that we can take $C_n \subset \cup_{i = 1}^nX_i$.
Intuitively, $C_{\epsilon}$ is a decent approximation of $E$. Let's make this precise:
\begin{align}
\mu^*(C_{\epsilon} \cap E^c) = & \mu^*\Big(\bigcup_{n = 1}^{\infty} (C_n \cap E^c)\Big) = \mu^*\Big(\bigcup_{n = 1}^{\infty} (C_n \cap E_n^c)\Big) \\ \le & \sum_{n = 1}^{\infty}\mu^*(C_n \setminus E_n) \le \epsilon.
\end{align}
(good job considering the $2^{-n}$, that really came in handy!)
Let's go back to our intuition: $C_{\epsilon}$ is an $\epsilon$-good approximation of $E$, therefore we would like to send $\epsilon$ to $0$ to get the best possible approximation. Since we want to end up in $\mathcal{A}_{\sigma \delta}$ what we need to do is to "discretize" the sets $C_{\epsilon}$ considering instead the sets $C_{\frac{1}{n}}$. (the notation is unfortunate, these resemble too much the $C_n$ sets, hopefully this won't cause any confusion)
Then clearly $$C := \bigcap_{n = 1}^{\infty}C_{\frac{1}{n}} \in \mathcal{A}_{\sigma \delta}$$ and satisfies $\mu^*(C \setminus E) = 0$, proving the claim.
To show this last equality notice that $$\mu^*(C \setminus E) \le \mu^*(C_{\frac{1}{n}} \setminus E) \le \frac{1}{n} \to 0.$$
Let me know if there is anything that I need to clarify!
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
Let $F \subset N$, where $N$ is a measurable null set, i.e. $N \in \mathcal{M}$ and $\mu(N) = 0$. We want to prove that $F \in \mathcal{M}^*$.
Using part $(b)$ and $(c)$ of the problem we solved yesterday, it is enough to show that there is $B \in \mathcal{A}_{\sigma \delta}$ such that $F \subset B$ and $\mu^*(B \setminus F) = 0$.
Since $\mathcal{M} \subset \mathcal{M}^*$, we have that $N \in \mathcal{M}^*$ and hence there is $C \in \mathcal{A}_{\sigma \delta}$ such that $N \subset C$ and $\mu^*(C \setminus N) = 0$. But this clearly implies that $$\mu^*(C \setminus F) \le \mu^*(C \setminus N) + \mu^*(N \setminus F) \le \mu^*(C \setminus N) + \mu^*(N) = 0 + \mu(N) = 0.$$
This should work, let me know what you think!