Step 1: Set the goal.
A reaction runs on its own (is spontaneous) when the cell potential is positive, $E^\circ_{\text{cell}} > 0$. We test each option with this single criterion.
Step 2: List the reduction potentials.
$E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}$, $\;E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44\text{ V}$, $\;E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.23\text{ V}$. A higher (less negative) value means a stronger pull to get reduced.
Step 3: Recall the working formula.
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] where the species being reduced is the cathode and the one being oxidised is the anode.
Step 4: Test option (A).
Here $\text{Ni}^{2+}$ is reduced and $\text{Fe}$ is oxidised: \[ E^\circ_{\text{cell}} = (-0.23) - (-0.44) = +0.21\text{ V} \] This is positive, so the reaction is spontaneous.
Step 5: Quickly rule out the rest.
In (B), (C), and (D) the metal that wants reduction least is forced to be reduced, so each gives a negative $E^\circ_{\text{cell}}$. For instance (C) gives $(-0.76) - (-0.44) = -0.32\text{ V}$, clearly non-spontaneous.
Step 6: Conclude.
Only option (A) yields a positive cell potential, so it is the spontaneous reaction.
\[ \boxed{\text{Ni}^{2+} + \text{Fe} \rightarrow \text{Ni} + \text{Fe}^{2+}} \]