The standard Gibbs energy for the given cell reaction in kJ mol$$^{-1}$$ at 298 K is:
Zn(s) + Cu$$^{2+}$$(aq) $$\rightarrow$$ Zn$$^{2+}$$(aq) + Cu(s),
E$$^0$$ = 2 V at 298 K
(Faraday's constant, F = 96000 C mol$$^{-1}$$)

We have to connect the standard cell potential with the standard Gibbs energy change. The thermodynamic relation that links these two quantities is stated as:

$$\Delta G^{\circ} = -\,n\,F\,E^{\circ}$$

Here,

$$n$$ is the number of electrons transferred in the balanced redox reaction,

$$F$$ is Faraday’s constant $$\bigl(F = 96000 \text{ C mol}^{-1}\bigr)$$,

$$E^{\circ}$$ is the standard cell potential.

Now we first determine $$n$$ for the given cell reaction:

$$\text{Zn(s)} \;+\; \text{Cu}^{2+}(aq) \;\longrightarrow\; \text{Zn}^{2+}(aq) \;+\; \text{Cu(s)}$$

In this reaction, zinc is oxidised from the 0 oxidation state to $$+2$$, therefore it loses two electrons. Simultaneously, copper(II) ion is reduced from $$+2$$ to $$0$$, thus it gains those two electrons. Hence, the total number of electrons transferred is

$$n = 2$$

Substituting the known values into the formula:

$$\Delta G^{\circ} = -\,\bigl(2\bigr)\,\bigl(96000\text{ C mol}^{-1}\bigr)\,\bigl(2\text{ V}\bigr)$$

We multiply step by step. First the constants:

$$2 \times 96000 = 192000$$

and then include the potential:

$$192000 \times 2 = 384000$$

Applying the negative sign from the formula, we obtain

$$\Delta G^{\circ} = -\,384000 \text{ J mol}^{-1}$$

Now we convert joules to kilojoules because the options are expressed in kJ mol$$^{-1}$$:

$$\Delta G^{\circ} = \frac{-384000}{1000} = -384 \text{ kJ mol}^{-1}$$

So the standard Gibbs energy change for the given cell reaction at 298 K is $$-384 \text{ kJ mol}^{-1}$$.

Hence, the correct answer is Option D.