Denote: $\bullet\;$ brick of type A: with dimensions $2\times5\times8$; $\bullet\;$ brick of type B: with dimensions $2\times3\times7$. The box contains $10\cdot11\cdot14=1540$ unit cubes; a brick of type A contains $2\cdot5\cdot8=80$ unit cubes; a brick of type B contains $2\cdot3\cdot7=42$ unit cubes. The possible number of bricks is a non-negative integer solution of the equation: $80x+42y=1540\Longrightarrow 21y=10(77-4x)\Longrightarrow y=10m$, where $m\in\mathbb{Z}\Longrightarrow$ $\Longrightarrow 21m=77-4x\Longrightarrow x=\dfrac{77-21m}{4}=19-5m-\dfrac{m-1}{4}\in\mathbb{Z}\Longrightarrow$ $\Longrightarrow \dfrac{m-1}{4}=n\in\mathbb{Z}\Longrightarrow m=4n+1$. Results: $x=14-21n\;;y=40n+10$. $x\ge0\;;y\ge0\Longrightarrow n=0\Longrightarrow x=14\;;y=10$. Hence, the equation has an unique non-negative integer solution: $(x,y)=(14,10)$ and $x+y=24$. Remains to prove: we can fill the box with a configuration of $14$ bricks A and $10$ bricks B. The solution in plane: we can fill a $7\times11$ rectangle with: a rectangle $2\times8$, a rectangle $5\times8$ and a rectangle $3\times7$. Results: we can fill a $11\times14$ rectangle with: $2$ rectangles $2\times8$, $2$ rectangles $5\times8$ and $2$ rectangles $3\times7$. The solution in space: The base of the box is the rectangle $11\times14$ and the height is $10$. We can put in the box: $2$ columns, each containing $2$ bricks A with the base $2\times8$ and the height $5$; $2$ columns, each containing $5$ bricks A with the base $5\times8$ and the height $2$; $2$ columns, each containing $5$ bricks B with the base $3\times7$ and the height $2$. Final answer: the total number of bricks is $24$, respectively $14$ bricks with the dimensions $2\times5\times8$ and $10$ bricks with the dimensions $2\times3\times7$.

The only possible number of bricks is $24,$ with $10$ 2x3x7 bricks and $14$ 2x5x8 bricks. We first show this is achievable, and then prove it is the unique solution. Instead of thinking of them as $14$ 2x5x8 bricks and $10$ 2x3x7 bricks, we can stack 7 bricks on top of each other and 2 bricks on top of each other, respectively, to get that we have two "blocks" of $5x8x14$ and $5$ "blocks" of $2x3x14$. Orient these bricks so that the $14$ edge is parallel to the z-axis. Then, as long as we can prove that there is a way to title a $10x11$ grid using 2 $5x8$ tiles and 5 $2x3$ tiles, we are done showing it is achievable. This is easy to see as we can just put two $5x8$ bricks so that the two "5" sides are adjacent to the side of $10,$ and this leaves an uncovered $10x3$ rectangle, which we can fill by arranging the five 2x3 bricks in a row with the 2 base horizontal. Now we show this the unique solution. Notice that for some number of bricks to work, the volume of the bricks and the volume of the large box must be the same. Suppose we have $a$ 2x5x8 bricks and $b$ 2x3x7 bricks. Then $80a+42b=10\cdot 11\cdot 14$. It is easy to see that $10|b$ and $7|a$. Letting $b=10y$ and $a=7x,$ we have that $4x+3y=11,$ and the only solution to this is $(x,y)=(2,1).$ Therefore, the only solution is $a=14$ and $b=10$. We are done.