Step 1: Recall the Nernst equation.
For a cell at room temperature, $E_{cell} = E^{\circ}_{cell} - \dfrac{0.06}{n}\log Q$, where $n$ is the number of electrons and $Q$ is the reaction quotient. Here we are given $\dfrac{2.303RT}{F} = 0.06$ V.
Step 2: Write the cell reaction and find n.
Iron is the anode and copper is the cathode, so \[ Fe + Cu^{2+} \rightarrow Fe^{2+} + Cu \] Two electrons are transferred, so $n = 2$.
Step 3: Write the reaction quotient Q.
Only ions appear in $Q$, so \[ Q = \frac{[Fe^{2+}]}{[Cu^{2+}]} = \frac{0.001}{0.1} = 10^{-2} \]
Step 4: Plug values into the Nernst equation.
With $\dfrac{0.06}{2} = 0.03$, \[ 0.82 = E^{\circ}_{cell} - 0.03 \log(10^{-2}) \]
Step 5: Evaluate the log term.
Since $\log(10^{-2}) = -2$, the equation becomes \[ 0.82 = E^{\circ}_{cell} - 0.03(-2) = E^{\circ}_{cell} + 0.06 \]
Step 6: Solve for the standard potential.
So $E^{\circ}_{cell} = 0.82 - 0.06 = 0.76$ V.
\[ \boxed{0.76 \text{ V}} \]