1. Write the cell reaction

For the Daniell cell:

$Zn|Zn^{2+}(1M)||Cu^{2+}(1M)|Cu$

The overall reaction is:

$Zn(s)+Cu^{2+}(aq) \rightarrow Zn^{2+}(aq)+Cu(s)$

Here, number of electrons transferred is

$n=2$

2. Condition for complete discharge

When the cell is completely discharged, it can no longer produce emf, so:

$E_{cell}=0$

Use the Nernst equation:

$E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q$

At $298\,K$ and for this reaction,

$Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$

So at complete discharge:

0=1.10-\frac{0.0591}{2}\log\left(\frac{[Zn^{2+}]}{[Cu^{2+}]}}\right)

Thus,

\frac{0.0591}{2}\log\left(\frac{[Zn^{2+}]}{[Cu^{2+}]}}\right)=1.10

\log\left(\frac{[Zn^{2+}]}{[Cu^{2+}]}}\right)=\frac{2\times 1.10}{0.0591}

\log\left(\frac{[Zn^{2+}]}{[Cu^{2+}]}}\right)\approx 37.3

Therefore,

\frac{[Zn^{2+}]}{[Cu^{2+}]}}=\text{antilog}(37.3)=10^{37.3}

3. Match with options

Option C is written as $10^{37.3}$ (the formatting in the options seems compressed as 1037.3).

So the correct option is:

$\boxed{C}$

4. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They agree.