Blocks...

We call the values in the three empty blocks in the bottom row $\alpha, \beta$ and $\gamma$ . Then we add our way up the pyramid as shown below:

Then from the pairs of blocks beneath the three red blocks in the top and middle rows, we get the equations

$\begin{array}{rlcrlccl} 120 + 2\alpha + 4\beta + 2\gamma &= \ \ 240 \ \ & \implies &\ \ \alpha + 2\beta + \gamma &= \ \ 60 & \small (1) \\ 12 + 2\alpha + \beta &= \ \ 67 & \implies & 2\alpha + \beta &= \ \ 55 & \small (2) \\ \beta + 2\gamma + 18 &= \ \ 53 & \implies & \beta + 2\gamma &= \ \ 35 & \small (3) \\ \\ & & & \alpha + \beta + \gamma &= \ \ 45 & \small (4) & &\small \text{[ Half of sum } (2) + (3) \ ] \\ \\ & & & \beta &= \ \ \boxed{15} & & &\small [ \ (1) - (4) \ ] \end{array}$

In figure 1, let $x$ be the unknown number on the second row leftmost, we observed that $a=x-67$ and $a=240-x-53=187-x$ . Then, equate the two equations

$a=a$

$x-67=187-x$

$2x=254$

$x=127$

It follows that $240-x=240-127=113$ and $a=x-67=127-67=60$

Fill up the other boxes as shown in figure 2.

In figure 3, let $y$ be the unknown number on the fourth row leftmost.

In figure 4, let $m$ be the number on the green box. We observed that $m=67-y-(y-12)=67-y-y+12=79-2y$ and $m=y-7-(42-y)=y-7-42+y=2y-49$ . Equate the two equations

$m=m$

$79-2y=2y-49$

$79+49=4y$

$y=32$

It follows that,

$m=79-2(32)=\boxed{15}$