A Daniell-type cell Zn(s) | Zn^2+(aq) \| Cu^2+(aq) | Cu(s) has E^°cell=1.10 V at 298 K. Take (2.303RT)/(F)=0.059 V.

The cell reaction is $\mathrm{Zn+Cu^{2+}\to Zn^{2+}+Cu}$ with $n=2$, so $E=E^\circ-\dfrac{0.059}{2}\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}$. Option A: $\log(1/0.01)=2$, so $E=1.10-0.059=1.041\ \mathrm{V}$, correct. Option B: $\log(0.1/1)=-1$, so $E=1.10+0.0295=1.1295\ \mathrm{V}$, which is higher than $1.10\ \mathrm{V}$, so B is wrong. Option C: $\log K=\dfrac{nE^\circ}{0.059}=\dfrac{2\times1.10}{0.059}=37.3$, so $K\approx2\times10^{37}$, which is outside the stated range, so C is wrong. Option D: emf is an intensive property fixed by $E^\circ$ and the concentrations, and electrode area changes only the current capacity, so D is wrong. Only A is correct. As a further check on the structure of the problem, remember that the Nernst correction depends only on the reaction quotient $Q=[\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}]$, so any change that lowers $Q$ raises the emf, and the emf falls to zero only when $Q$ equals $K$, that is, at equilibrium.