The relationship between standard cell potential ($E^\circ_{\text{cell}}$) and equilibrium constant ($K_c$) at $298 \text{ K}$ is given by: $E^\circ_{\text{cell}} = \frac{0.059}{n} \log K_c$ For the given reaction: $\text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$ The number of electrons transferred ($n$) is $2$. Given $E^\circ = 0.46 \text{ V}$, we substitute the values into the equation: $0.46 = \frac{0.059}{2} \log K_c$ $\log K_c = \frac{0.46 \times 2}{0.059} = \frac{0.92}{0.059} = 15.59 \approx 15.6$ Taking the antilog of both sides: $K_c = \text{antilog}(15.6) = 3.92 \times 10^{15}$ Rounding off to the nearest option, we get $4.0 \times 10^{15}$.