Consider the above electrochemical cell where a metal electrode (M)...

The given setup is a concentration cell with identical electrodes (M) and different electrolyte concentrations ($c_1$ and $c_2$).
For a concentration cell, the standard cell potential $E^0_{cell} = 0$.
The cell reactions are:
Anode: $M(s) \rightarrow M^+(c_{anode}) + e^-$
Cathode: $M^+(c_{cathode}) + e^- \rightarrow M(s)$
Net reaction: $M^+(c_{cathode}) \rightarrow M^+(c_{anode})$
Using the Nernst equation at $298\text{ K}$:
$E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{c_{anode}}{c_{cathode}}$
$E_{cell} = 0 - 0.0591 \log \frac{c_{anode}}{c_{cathode}}$
For a positive cell potential ($E_{cell} > 0$), the term $\log \frac{c_{anode}}{c_{cathode}}$ must be negative.
This requires $\frac{c_{anode}}{c_{cathode}} < 1$, which means $c_{cathode} > c_{anode}$.
If $c_1$ is present at the cathode and $c_2$ is at the anode, then for $E_{cell} > 0$, we must have $c_1 > c_2$.
Checking the options:
(1)$If c_1$ is at cathode, $c_1 < c_2$ (Incorrect)
(2)$If c_1$ is at anode, $c_1 > c_2$ (Incorrect, would require $c_2 > c_1$)
(3)$If c_1$ is at cathode, $c_1 > c_2$ (Correct)
(4)$If c_1$ is at anode, $c_1 = c_2$ (Incorrect, $E_{cell}$ would be zero)