Build it tall

The volume (and thus the weight) goes as $l^3$ , and the structural strength goes as $l^2$ . Therefore, the strength to weight ratio goes as $l^{-1}$ . Since the human-scale building is 30 times as big, it is only $1/30$ as strong relative to its weight. Therefore, the expected number of floors the human-scale building will support is:

$N = \frac{500}{30} = 16.67 \approx 15$

The weight of each cube floor in the human scale building is, in kilograms

$3 \cdot 3 \cdot 3 \cdot 0.3 \cdot 340=2754$

Scaling up one cube's ability to sustain a load in the human scale building is, in kilograms

$\dfrac{3}{0.1} \cdot \dfrac{3}{0.1} \cdot 50 = 45000$

The ratio is is approximately $16.34$ , so $15$ stories

Wrong way to do this:

$\dfrac{3}{0.1} \cdot \dfrac{3}{0.1} \cdot \dfrac{3}{0.1} \cdot 50 = 1350000$

giving us a ratio of approximately $490.20$ , or optimistically $500$ stories

The compression strength of any material is a function of cross section area, not volume. Here, we disregard structural failure from buckling, as that does depend on the height of the tower relative to its width.

Additional comments:

The cross section area of the model cube, assuming that there is a cubical void $0.7$ of the total volume, is, in square centimeters

$10 \cdot 10 \cdot \ (1-(0.7)^\frac{2}{3}) = 21.16$ or about $3.5$ square inches. If this sustains a load of $50$ pounds, this works out to roughly $14$ pounds per square inch. Steels commonly can safely sustain loads of $10$ s of thousands of pounds per square inch. The strength of this model cube would be typical for cardboard boxes of about the same size. Some homes and buildings have successfully been constructed using laminated cardboard, but they are generally limited to one story.

Let $W_0$ be the weight the model floor can sustain, $n$ , the number of floors the human-scale building can sustain and $W_s = (1-f)\rho l^3g$ , the weight of a single story. Assuming the human-scale floor can sustain the same pressure as the model floor, we have: $P = \frac{W_0}{l_0^2} = \frac{nW_s}{l^2}$ .

So, $n = \left(\frac{l}{l_0}\right)^2 \frac{W_0}{W_s}$ .

The ratio $l:l_0$ is 30:1, the model floor can sustain 50 kg and a single story weighs 2754 kg. The equation finally results in:

$n = 900 * \frac{50}{2754} = 16.34 \simeq 15$

The linear dimension of the human-scale structure is 30 times bigger (3m/0.1) than that of the model, so the area of the floor is 900 ( $l^2$ ) times bigger and the volume 27000 ( $l^3$ ) times. When the structure gets bigger in area, weight and area cancel each other out since while there is ( $\frac{l_1}{l_0}$ ) $^2$ times more weight there is also ( $\frac{l_1}{l_0}$ ) $^2$ times more area to distribute the weight.

This doesn't happen with height, an increase in height does not come with more capacity to support the increase in weight. Therefore, when we increase the volume of the building by $30^3$ it only matters the factor of height (30). Since each floor is 30 times taller each has to support 30 times more weight, so the structure will only support $1/30$ of the stories compared to the model. The model can support 500 times its weight so 500/30 =16.67