Think of this problem as: how many independent numbers do you need to fully describe the stiffness of the "worst case" elastic material, one with no special symmetry at all (fully anisotropic, also called triclinic)?
- 36: this is what you would get if you just multiplied the 6 stress components by the 6 strain components without using any physical restriction. It overcounts the real material behavior.
- 21: strain energy stored in an elastic solid depends only on the current state of strain, not on the order in which the strains were applied. This "path independence" forces the stiffness matrix relating the 6 stress and 6 strain components to be symmetric. A symmetric $6\times6$ matrix has 6 numbers on its diagonal and 15 more above the diagonal (the ones below just repeat them), for $6+15=21$ truly independent numbers.
- 10: this count is too low for a general anisotropic solid. Reduced counts like this only show up once you assume extra material symmetry (such as one or more planes of symmetry).
- 2: this is the count for an isotropic material (same stiffness in every direction), described fully by just two constants, for example Young's modulus and Poisson's ratio. That is the most symmetric case, the exact opposite of "fully anisotropic".
Since the question asks about a fully anisotropic material with no symmetry assumed, the answer is the largest physically allowed count, 21.
Let's summarize:
- Symmetry of the stiffness matrix (from strain energy) always cuts 36 down to 21, for any elastic solid.
- Extra material symmetry (orthotropic, isotropic, etc.) cuts the 21 down further, it does not raise it.
So a fully anisotropic linear elastic material has 21 independent elastic constants, option (B).