1. We need the dimensions of $\frac{B^2}{2\mu_0}$ Since $2$ is dimensionless, this is the same as finding dimensions of $\frac{B^2}{\mu_0}.$

2. Use the known physical meaning: $\frac{B^2}{2\mu_0}$ is the magnetic energy density.

Energy density $=$ energy per unit volume.

So its dimensions are: $\frac{[\text{Energy}]}{[\text{Volume}]}=\frac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}.$

3. Therefore, $\left[\frac{B^2}{2\mu_0}\right]=ML^{-1}T^{-2}.$

4. Match with the options:

• A: $ML^2T^{-2}$
• B: $MLT^{-2}$
• C: $ML^{-1}T^{-2}$
• D: $ML^2T^{-1}$

Hence, the correct option is: $\boxed{C}$

5. Verification with stored answer: Stored correct answer = C, which matches our result.