The Stefan-Boltzmann constant $\sigma_{S}$ has dimensions $[M T^{-3} K^{-4}]$. The Boltzmann constant $k_{B}$ has dimensions $[M L^2 T^{-2} K^{-1}]$. The Wien's displacement law constant $b$ has dimensions $[L K]$.
We need to find the dimensions of the product $\sigma_{S} \cdot k_{B}^{-1} \cdot b$. Substituting the dimensions:
$$[Dimension] = [M T^{-3} K^{-4}] \cdot [M L^2 T^{-2} K^{-1}]^{-1} \cdot [L K]$$First, invert the dimensions of $k_{B}$:
$$[k_{B}^{-1}] = [M^{-1} L^{-2} T^2 K^1]$$Now multiply all terms:
$$[Dimension] = [M T^{-3} K^{-4}] \cdot [M^{-1} L^{-2} T^2 K^1] \cdot [L K]$$ $$[Dimension] = [M^{1-1}] \cdot [L^{-2+1}] \cdot [T^{-3+2}] \cdot [K^{-4+1+1}]$$ $$[Dimension] = [M^0] \cdot [L^{-1}] \cdot [T^{-1}] \cdot [K^{-2}]$$Final Answer: [L^{-1}T^{-1}K^{-2}]
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