1. Write the cell reaction and identify $n$

Given reaction: $Cu_{(s)} + 2Ag^+(10^{-3}M) \to Cu^{2+}(0.250M) + 2Ag_{(s)}$

Here, copper is oxidized: $Cu \to Cu^{2+} + 2e^-$

Silver ion is reduced: $2Ag^+ + 2e^- \to 2Ag$

So, the number of electrons transferred is: $n=2$

2. Use the Nernst equation

For the reaction, $E_{cell} = E_{cell}^\circ - \frac{0.0591}{n}\log Q$

At $298\,K$, with $n=2$: $E_{cell} = 2.97 - \frac{0.0591}{2}\log Q$

3. Calculate the reaction quotient $Q$

For $Cu_{(s)} + 2Ag^+ \to Cu^{2+} + 2Ag_{(s)}$

Solids are omitted, so $Q = \frac{[Cu^{2+}]}{[Ag^+]^2}$

Substitute the given concentrations: $Q = \frac{0.250}{(10^{-3})^2} = 0.250 \times 10^6 = 2.5 \times 10^5$

4. Evaluate $\log Q$

$\log(2.5 \times 10^5) = \log 2.5 + 5 = 0.3979 + 5 = 5.3979$

5. Substitute into Nernst equation

$E_{cell} = 2.97 - \frac{0.0591}{2}(5.3979)$

$\frac{0.0591}{2} = 0.02955$

So, $E_{cell} = 2.97 - 0.02955 \times 5.3979$

$0.02955 \times 5.3979 \approx 0.1595$

Thus, $E_{cell} \approx 2.97 - 0.1595 = 2.8105\,V$

6. Nearest integer

$E_{cell} \approx 3\,V$

Therefore, the required nearest integer is 3.