Explanation:

Let the initial volume of the solution be $V_1 = 6 \text{ litres}$. **Initial Composition:**\nWater content = $30\%$.\nAlkali content = $70\%$. \nCalculate the initial amount of water and alkali:\nAmount of water = $30\%$ of $6 \text{ litres} = 0.30 \times 6 = 1.8 \text{ litres}$.\nAmount of alkali = $70\%$ of $6 \text{ litres} = 0.70 \times 6 = 4.2 \text{ litres}$. (Check: $1.8 + 4.2 = 6 \text{ litres}$, which is the total initial volume). **Adding Water:**\nLet $x$ be the quantity of water added (in litres).\nCrucially, the amount of alkali in the solution remains constant, as only water is being added.\nNew total volume of solution ($V_2$) = $(6 + x) \text{ litres}$.\nNew amount of water = $(1.8 + x) \text{ litres}$.\nAmount of alkali (constant) = $4.2 \text{ litres}$. **Final Composition:**\nIn the new solution, the water content is $40\%$.\nThis means the alkali content is $100\% - 40\% = 60\%$. \nNow, we can set up an equation based on the constant amount of alkali: $60\%$ of $V_2 = \text{Amount of alkali}$ $0.60 \times (6 + x) = 4.2$ \nSolve for $x$: $6 + x = \frac{4.2}{0.60}$ $6 + x = \frac{420}{60}$ $6 + x = 7$ $x = 7 - 6$ $x = 1 \text{ litre}$. \nThe quantity of water to be added is $1$ litre. Since $1 \text{ litre} = 1000 \text{ ml}$, the answer is $1000 \text{ ml}$. \nLet's evaluate the options:\nOption (1) $500 \text{ ml}$ ($0.5 \text{ litres}$): If $0.5 \text{ litres}$ were added, the new water percentage would be $\frac{1.8+0.5}{6+0.5} = \frac{2.3}{6.5} \approx 35.38\%$, not $40\%$.\nOption (2) $700 \text{ ml}$ ($0.7 \text{ litres}$): If $0.7 \text{ litres}$ were added, the new water percentage would be $\frac{1.8+0.7}{6+0.7} = \frac{2.5}{6.7} \approx 37.31\%$, not $40\%$.\nOption (3) $900 \text{ ml}$ ($0.9 \text{ litres}$): If $0.9 \text{ litres}$ were added, the new water percentage would be $\frac{1.8+0.9}{6+0.9} = \frac{2.7}{6.9} \approx 39.13\%$, not $40\%$.\nOption (4) $1 \text{ litre}$: This matches our calculated value. If $1 \text{ litre}$ is added, the new water percentage is $\frac{1.8+1}{6+1} = \frac{2.8}{7} = 0.4 = 40\%$. This is correct. \nTherefore, $1$ litre of water should be added. The correct option is (4) 1 litre.