In three containers - yellow, white and black, $120$ marbles are placed. There are twice as many marbles in the white and yellow containers combined as there are in the black container. Also, there are twice as many marbles in the black container as there are in the yellow container. How many marbles are there in the white container?

Let the number of marbles in the yellow, white, and black containers be $Y$, $W$, and $B$ respectively.
According to the problem:

1. Total number of marbles:

$$Y+W+B=120$$

2. White and yellow combined have twice as many as black:

$$W+Y=2B$$

3. Black has twice as many as yellow:

$$B=2Y⟹Y=\frac{B}{2}$$

Substitute $Y=\frac{B}{2}$ into the second equation:

$$W+\frac{B}{2}=2B⟹W=2B-\frac{B}{2}=\frac{3B}{2}$$

Now substitute the expressions for $Y$ and $W$ into the total marbles equation:

$$\frac{B}{2}+\frac{3B}{2}+B=120$$

$$\frac{4B}{2}+B=120$$

$$2B+B=120$$

$$3B=120⟹B=40$$

Find the number of marbles in the white container ($W$):

$$W=\frac{3\times 40}{2}=60$$