1. Given expression

   $u=\frac{\alpha}{\beta}\sin\left(\frac{\alpha x}{kt}\right)$

   Here, $u$ is energy density.

2. Dimension of energy density

   Energy density $=$ energy per unit volume.

   $[u]=\frac{[\text{energy}]}{[\text{volume}]}=\frac{[ML^2T^{-2}]}{[L^3]}=[ML^{-1}T^{-2}]$

3. Use the fact that argument of sine is dimensionless

   Since $\sin\left(\frac{\alpha x}{kt}\right)$ must have a dimensionless argument,

   $\left[\frac{\alpha x}{kt}\right]=1$

   Therefore, $[\alpha][x]=[k][t]$

4. Dimensions of $k t$

   Boltzmann constant $k$ has dimension:

   $[k]=\frac{[\text{energy}]}{[\theta]}=[ML^2T^{-2}\theta^{-1}]$

   Temperature $t$ has dimension $[\theta]$.

   Hence, $[kt]=[ML^2T^{-2}\theta^{-1}][\theta]=[ML^2T^{-2}]$

5. Find dimension of $\alpha$

   Since $[x]=[L]$,

   $[\alpha]=\frac{[kt]}{[x]}=\frac{[ML^2T^{-2}]}{[L]}=[MLT^{-2}]$

6. Now use prefactor $\alpha/\beta$

   Since sine is dimensionless,

   $\left[\frac{\alpha}{\beta}\right]=[u]=[ML^{-1}T^{-2}]$

   Therefore,

   $[\beta]=\frac{[\alpha]}{[u]}=\frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]}=[L^2]$

   So,

   $[\beta]=[M^0L^2T^0]$

7. Match with options

   This corresponds to Option D.

Comparison with stored answer: Stored correct answer is D, which matches our derived answer.