1. Write the cell reaction

For the cell $\mathrm{Mg(s)|Mg^{2+}(aq)\parallel Cu^{2+}(aq)|Cu(s)}$

Anode: $\mathrm{Mg(s) \to Mg^{2+}(aq) + 2e^-}$ Cathode: $\mathrm{Cu^{2+}(aq) + 2e^- \to Cu(s)}$

Overall reaction: $\mathrm{Mg(s) + Cu^{2+}(aq) \to Mg^{2+}(aq) + Cu(s)}$

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

2. Use the Nernst equation

For the reaction, $Q=\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Cu^{2+}}]}$

Initially, both concentrations are $1\,M$, so $Q=1$ and hence $E^\circ = 2.70\,V$

When only $[\mathrm{Mg^{2+}}]$ is changed to $x\,M$ and $[\mathrm{Cu^{2+}}]=1\,M$, $Q=x$

Then, $E = E^\circ - \frac{RT}{nF}\ln Q$ $2.67 = 2.70 - \frac{RT}{2F}\ln x$

3. Substitute the given data

Rearranging, $2.70 - 2.67 = \frac{RT}{2F}\ln x$ $0.03 = \frac{RT}{2F}\ln x$

Given: $\frac{F}{R}=11500\,K\,V^{-1} \implies \frac{R}{F}=\frac{1}{11500}$

At $T=300\,K$, $\frac{RT}{F}=\frac{300}{11500}$

So, $0.03 = \frac{1}{2}\cdot \frac{300}{11500}\ln x$

$0.03 = \frac{150}{11500}\ln x$

$\ln x = 0.03\cdot \frac{11500}{150}$

$\ln x = \frac{345}{150} = 2.30$

Using the given fact $\ln 10 = 2.30$, $x=10$

4. Final answer

$\boxed{10}$

The derived answer matches the stored correct answer.