The balanced chemical equation for the oxidation of chloride ions to chlorine gas at the anode is: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$ This indicates that the formation of $1 \text{ mole}$ of $\text{Cl}_2$ gas requires the passage of $2 \text{ moles}$ of electrons, which corresponds to a charge of $2 \text{ Faradays}$ ($2 \times 96500 \text{ C}$). For producing $0.10 \text{ mol}$ of $\text{Cl}_2$, the required charge ($Q$) is: $Q = 0.10 \times 2 \times 96500 \text{ C} = 19300 \text{ C}$ We know that charge $Q = I \times t$, where $I$ is the current and $t$ is the time in seconds. Given current $I = 3 \text{ A}$, $t = \frac{Q}{I} = \frac{19300}{3} \text{ s} = 6433.33 \text{ s}$ To convert the time into minutes, we divide by $60$: $t \text{ (in minutes)} = \frac{6433.33}{60} \approx 107.22 \text{ minutes}$ Since $107.22 \text{ minutes}$ is closest to $110 \text{ minutes}$ among the given options, $110 \text{ minutes}$ is the appropriate answer.