For the given cell; $$\text{Cu}(s)|\text{Cu}^{2+}(C_1\,M)||\text{Cu}^{2+}(C_2\,M)|\text{Cu}(s)$$
change in Gibbs energy $$(\Delta G)$$ is negative, it:

We start by examining the given cell notation $$$\text{Cu}(s)\;|\;\text{Cu}^{2+}(C_1\,\text{M})\;||\;\text{Cu}^{2+}(C_2\,\text{M})\;|\;\text{Cu}(s)$$$ carefully. In the conventional cell diagram the species written on the left form the anode (oxidation takes place), while those on the right form the cathode (reduction takes place).

Thus, at the anode we have the oxidation half-reaction

$$\text{Cu}(s)\;\longrightarrow\;\text{Cu}^{2+}(C_1)\;+\;2e^-$$

and at the cathode the reduction half-reaction

$$\text{Cu}^{2+}(C_2)\;+\;2e^-\;\longrightarrow\;\text{Cu}(s).$$

Adding the two half-reactions gives the overall cell reaction

$$\text{Cu}^{2+}(C_2)\;\longrightarrow\;\text{Cu}^{2+}(C_1).$$

Notice that solid copper cancels out and only the two ionic concentrations appear. For such a concentration cell the standard e.m.f. is zero, $$E^\circ_{\text{cell}} = 0,$$ because both electrodes are made of the same material.

Now we invoke the Nernst equation. First we state it in its general form:

$$E = E^\circ_{\text{cell}} - \frac{0.0591}{n}\,\log Q\qquad\text{at }T = 298\text{ K},$$

where $$Q$$ is the reaction quotient and $$n$$ is the number of electrons transferred.

Here, $$n = 2$$ (two electrons in each half-reaction). The reaction quotient for the overall reaction

$$\text{Cu}^{2+}(C_2)\;\longrightarrow\;\text{Cu}^{2+}(C_1)$$

is simply

$$Q = \frac{[\text{Cu}^{2+}]_{\text{products}}}{[\text{Cu}^{2+}]_{\text{reactants}}} =\frac{C_1}{C_2}.$$

Substituting $$E^\circ_{\text{cell}} = 0$$, $$n = 2$$ and $$Q = \dfrac{C_1}{C_2}$$ in the Nernst equation, we obtain

$$E \;=\; 0 \;-\;\frac{0.0591}{2}\,\log\!\left(\frac{C_1}{C_2}\right) \;=\;\frac{0.0591}{2}\,\log\!\left(\frac{C_2}{C_1}\right).$$

Next we relate the e.m.f. to the change in Gibbs free energy. We state the equation first:

$$\Delta G = -\,nF\,E,$$

where $$F$$ is the Faraday constant. The quantity $$F$$ is always positive, and $$n = 2$$ is also positive, so the sign of $$\Delta G$$ is completely determined by the sign of $$E$$.

For the process to be spontaneous we require $$\Delta G \lt 0,$$ which demands $$E \gt 0.$$

From the expression

$$E \;=\;\frac{0.0591}{2}\,\log\!\left(\frac{C_2}{C_1}\right),$$

we see that $$E \gt 0$$ if and only if the logarithm is positive, i.e.

$$\log\!\left(\frac{C_2}{C_1}\right) \gt 0 \;\;\Longrightarrow\;\;\frac{C_2}{C_1} \gt 1 \;\;\Longrightarrow\;\;C_2 \gt C_1.$$

Therefore the concentration on the cathode side must exceed that on the anode side for the cell to run spontaneously (giving negative $$\Delta G$$).

Examining the four options:

A. $$C_1 = C_2$$   gives $$C_2/C_1 = 1$$ and hence $$E = 0$$ → not spontaneous.

B. $$C_2 = \dfrac{C_1}{\sqrt{2}}$$   gives $$C_2/C_1 \lt 1$$ → $$E \lt 0$$ → not spontaneous.

C. $$C_1 = 2C_2$$   also yields $$C_2/C_1 \lt 1$$ → $$E \lt 0.$$

D. $$C_2 = \sqrt{2}\,C_1$$   gives $$\dfrac{C_2}{C_1} = \sqrt{2} \gt 1,$$ so

$$E = \frac{0.0591}{2}\,\log(\sqrt{2}) \gt 0 \qquad\Rightarrow\qquad \Delta G \lt 0.$$

Only Option D satisfies the requirement $$\Delta G \lt 0.$$

Hence, the correct answer is Option 4.