We will prove the following general statement:

Given a $2n \times 2n \times 2n$ box of $8n^3$ unit white cubes. An and Binh play a game with the box. An selects some bands of size $1 \times 1 \times n$ such that any two chosen bands have no common points and then changes all cells on these bands to black. Binh then selects some unit cubes and asks An what color of these cells. In this case, Binh needs to choose at least $6n^2$ in order to determine all black cells based on An's answer.

Let $S_n$ be the set of cells that Binh uses to ask An and for each chosen cell $u$, let $R_u$ be the union of three $1 \times 1 \times n$ bands (vertical, horizontal and diagonal ones) passing through $u$.

Since any two chosen bands have no common point, for any black cell $u$ there are at least two other chosen cells on $R_u$. Otherwise, Binh cannot determine exactly which band passing through $u$ is black. Now, we assign a tuple $(a,b,c)$ for each cell $u$ of the box as follows:
* $a=2$ if the horizontal band through $u$ has no more in $S_n$ and $a=1$ otherwise.
* $b=2$ if the vertical band through $u$ has no more in $S_n$ and $b=1$ otherwise.
* $c=2$ if the diagonal band through $u$ has no more in $S_n$ and $c=1$ otherwise.

From the above observation, two of $a,b,c$ are $1$ and the third one is at most $2$. This implies that $a+b+c \le 4$.
Let $T$ be the sum of all assigned numbers on the box, we have
$$T = \sum_{u \in S_n} (a+b+c) \le 4|S_n|.$$
On the other hand, there is at least one chosen cell on each $1 \times 1 \times n$ band (in any direction) of the box. This implies that $T \ge 2 \cdot 3(2n)^2 = 24n^2$.
Therefore, we have $4|S_n| \ge 24n^2$ or $|S_n| \ge 6n^2$.

We now have the following statements:
- In the $2 \times 2 \times 2$ box, Binh needs to choose at least $6$ cells.
- In the $10 \times 10 \times 10$ box, Binh needs to choose at least $150$ cells.

We first show the easiest case: $2 \times 2 \times 2$ box. In this case, Binh removes two opposite cells and asks for the color of the remaining ones. It is easy to check that these chosen cells work.

Now we can choose $150$ cells in the $10 \times 10 \times 10$ box as follows. We divide the box into $5$ layers of size $10 \times 10 \times 2$ and each layer into $25$ boxes of size $2 \times 2 \times 2$ and label each box as in the picture below.

In the ith layer ($i=1,2,3,4,5$), Binh picks the boxes labeled i and removes any two opposite cells in these boxes and asks for the color of the remaining ones. It is easy to see that if we project chosen cells into any face of the box, the shadow covers the whole face. This implies that any chosen $1 \times 1 \times n$ band of An passes through one of the chosen $2 \times 2 \times 2$ boxes of Binh and Binh now can easily determine whether this band is black or not.

Hence, the least number of cells that Binh must choose is $150$.