[Solution from source] σ = [M T⁻³ K⁻⁴]; k_B = [M L² T⁻² K⁻¹]; b = [L K]. σ k_B⁻¹ b = [M T⁻³ K⁻⁴][M⁻¹ L⁻² T² K][L K] = [L⁻¹ T⁻¹ K⁻²]. Answer (2).
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