A certain amount of water was poured into a 300-litre container and the remaining por tion of the container was filled with milk. Then an amount of this solution was taken out from the container, which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was:

The correct answer is 30.

Let the initial volume of water poured into the container be $x$ litres.
Since the container's total capacity is 300 litres, the remaining volume of the container was filled with milk.
Therefore, the initial volume of milk in the container is:
$300-x$ litres.

Since the container is fully filled, the total volume of the mixture is 300 litres.
The concentration of milk in this initial mixture is:
$\frac{300-x}{300}$

Next, an amount of this solution is taken out of the container. The volume of the mixture removed is twice the volume of the water initially poured, which is:
$2x$ litres.

Since the mixture is homogeneous, the fraction of milk in the removed mixture is the same as its concentration in the container.
The volume of milk removed from the container is:
$2x\times \left(\frac{300-x}{300}\right)$ litres.

After removing this volume of solution, the container is refilled with water to its full capacity again. Refilling with water does not add any milk. Thus, the volume of milk remaining in the container after this operation is:
$\text{Remaining Milk}=\left(300-x\right)-2x\left(\frac{300-x}{300}\right)$ litres.

We can factor out $\left(300-x\right)$ to simplify the expression for the remaining volume of milk:
$\text{Remaining Milk}=\left(300-x\right)\left(1-\frac{2x}{300}\right)$
$\text{Remaining Milk}=\left(300-x\right)\left(\frac{300-2x}{300}\right)$ litres.

The resulting solution has a total volume of 300 litres (since it was refilled) and contains 72% milk.
So, the volume of milk in the final solution is:
$72\%\text{ of }300=\frac{72}{100}\times 300=216$ litres.

Now, we equate the two expressions for the remaining volume of milk:
$\left(300-x\right)\left(\frac{300-2x}{300}\right)=216$

Multiply both sides by 300:
$\left(300-x\right)\left(300-2x\right)=216\times 300$
$\left(300-x\right)\left(300-2x\right)=64800$

Expand the left side of the equation:
$90000-600x-300x+2x^2=64800$
$2x^2-900x+90000-64800=0$
$2x^2-900x+25200=0$

Divide the entire equation by 2 to simplify:
$x^2-450x+12600=0$

Now, we solve this quadratic equation using factorization. We need two numbers that multiply to 12600 and add up to -450.
These numbers are -420 and -30, because:
$-420\times -30=12600$
$-420-30=-450$

Rewriting the equation:
$x^2-420x-30x+12600=0$
$x\left(x-420\right)-30\left(x-420\right)=0$
$\left(x-30\right)\left(x-420\right)=0$

This gives two possible values for $x$:
$x=30$ or $x=420$

Since the container's total capacity is 300 litres, the initial water volume $x$ must be less than 300 litres.
Therefore, $x=420$ is not possible.
Thus, the volume of water initially poured into the container was 30 litres.