Step 1: Note what the cell does.
In $2M^{3+} + 2I^{-} \rightarrow 2M^{2+} + I_2$, each $M^{3+}$ gains one electron and two of them react, so a total of $n = 2$ electrons are transferred. We must find $\log K_c$.
Step 2: Link the standard cell potential to the equilibrium constant.
At equilibrium the free energy gives \[ nFE^\circ_{cell} = RT \ln K_c = 2.303\,RT \log K_c \] This is the bridge between electrochemistry and equilibrium.
Step 3: Rearrange for $\log K_c$.
\[ \log K_c = \frac{nFE^\circ_{cell}}{2.303\,RT} \]
Step 4: Put in the given values.
Here $n=2$, $F=96500$, $E^\circ_{cell}=0.235$, $R=8.3$, $T=298$. The top becomes $2 \times 96500 \times 0.235 = 45355$.
Step 5: Work out the bottom.
$2.303 \times 8.3 \times 298 = 2.303 \times 2473.4 \approx 5696$.
Step 6: Divide and conclude.
\[ \log K_c = \frac{45355}{5696} \approx 7.96 \] So the answer is about $7.96$.
\[ \boxed{7.96} \]