The energy density u of a system is given by the relation u =...

The argument of the sine function must be dimensionless. Therefore,
$\left[\dfrac{\alpha x}{k_B T}\right] = [\mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^0]$

The term $k_B T$ represents thermal energy, so its dimensional formula is $[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]$.
Substituting the dimensions of distance $x$ as $[\mathrm{L}]$:
$[\alpha] = \dfrac{[k_B T]}{[x]} = \dfrac{[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]}{[\mathrm{L}]} = [\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]$

The quantity $u$ is energy density (energy per unit volume), so its dimensional formula is:
$[u] = \dfrac{[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]}{[\mathrm{L}^3]} = [\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}]$

From the given equation, the dimensions of $u$ must equal the dimensions of $\dfrac{\alpha}{\beta}$:
$[\beta] = \dfrac{[\alpha]}{[u]} = \dfrac{[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}]}{[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}]} = [\mathrm{L}^2]$

The SI base unit for $[\mathrm{L}^2]$ is $\mathrm{m}^2$.

Answer: $\mathrm{m}^2$