Step 1: Write the balanced overall cell reaction first.
The anode gives $A \to A^{+} + e^{-}$ and the cathode gives $B^{2+} + 2e^{-} \to B$. To balance electrons we need two A atoms for every one B, so the overall reaction is:
\[ 2A + B^{2+} \to 2A^{+} + B \]
This tells us the number of electrons transferred is $n = 2$, which we will need for the Nernst equation.
Step 2: Confirm which electrode is the cathode.
The electrode with the higher standard reduction potential is reduced, so $E^{\circ}_{B^{2+}/B} = 3\,V$ is the cathode and $E^{\circ}_{A^{+}/A} = 1\,V$ is the anode.
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 3 - 1 = 2\,V \]
Step 3: Apply the Nernst equation properly instead of assuming it has no effect.
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q \]
where $Q$ is the reaction quotient, written as products over reactants using the ions actually in the cell notation: $Q = \dfrac{[A^{+}]^2}{[B^{2+}]}$.
Step 4: Substitute the given concentrations.
\[ Q = \frac{(0.1)^2}{0.01} = \frac{0.01}{0.01} = 1 \]
Step 5: Use the fact that $\log 1 = 0$.
Since $Q = 1$ exactly, the correction term $\dfrac{0.0591}{n}\log Q$ becomes zero, so the concentration terms do not shift the answer away from the standard value at all.
\[ E_{cell} = 2 - \frac{0.0591}{2} \times 0 = 2\,V \]
Step 6: Final answer.
\[ \boxed{2.0\,V} \]