To determine how the stress in the leg changes when a man's linear dimensions increase by a factor of 9, we can analyze the relationship between the dimensions and the stress on the legs.

First, let's establish some relationships:

1. Linear dimensions (length, width, height) increase by a factor of 9.


2. Density remains the same.

Stress is defined as force per unit area:

$$ \text{Stress} = \dfrac{\text{Force}}{\text{Area}} $$

Since density remains constant, the volume and thus the mass of the man will change according to the cube of the linear dimensions. Since the linear dimension changes by a factor of 9, the volume (and hence the mass) changes by a factor of:

$$ 9^3 = 729 $$

Therefore, the weight (force due to gravity) also increases by a factor of 729.

The cross-sectional area of the leg is proportional to the square of the linear dimensions. Thus, if the linear dimension increases by a factor of 9, the cross-sectional area increases by a factor of:

$$ 9^2 = 81 $$

Now, substituting these factors into the stress formula:

$$ \text{New Stress} = \dfrac{\text{New Force}}{\text{New Area}} = \dfrac{729 \times \text{Original Force}}{81 \times \text{Original Area}} $$

Simplifying, we get:

$$ \text{New Stress} = 9 \times \dfrac{\text{Original Force}}{\text{Original Area}} = 9 \times \text{Original Stress} $$

Thus, the stress in the leg increases by a factor of 9. Therefore, the correct answer is:

Option B: 9