A

Step 1: Calculate the heat required to melt the ice. The latent heat of ice is given as $L_{ice} = 3.4 \times 10^5 \text{ J/kg}$. For 0.020 kg of ice, the heat required ($Q_{melt}$) is:
$Q_{melt} = m_{ice} \times L_{ice} = 0.020 \text{ kg} \times 3.4 \times 10^5 \text{ J/kg} = 6800 \text{ J}$.

Step 2: Calculate the heat released by the steam when it condenses. The specific latent heat of steam is $L_{steam} = 2.3 \times 10^6 \text{ J/kg}$. Let $m_{steam}$ be the mass of steam needed. Therefore, the heat released by the steam when it condenses is:
$Q_{steam} = m_{steam} \times L_{steam}$.

Step 3: Since all the ice melts, we have:
$Q_{steam} = Q_{melt}$. So, we set them equal:
$m_{steam} \times 2.3 \times 10^6 \text{ J/kg} = 6800 \text{ J}$.
Solving for $m_{steam}$ gives:
$m_{steam} = \frac{6800 \text{ J}}{2.3 \times 10^6 \text{ J/kg}} \approx 0.00296 \text{ kg}$.

Step 4: Now, calculate the total mass of water in the container after all the ice has melted. Initially, we have:
0.10 kg of water + 0.020 kg of ice (which becomes water) + m_{steam} (the condensed steam).
Thus, the total mass is:
$m_{water} = 0.10 \text{ kg} + 0.020 \text{ kg} + 0.00296 \text{ kg} \approx 0.12296 \text{ kg}$.

Hence, the total amount of water in the container is approximately 0.123 kg.