$B I=\sqrt{\left(\frac{\epsilon k_{ B } T}{n q^2}\right)}$

$D I=\sqrt{\left(\frac{q^2}{ en ^{1 / 3} k_B T}\right)}$

As given condition $\rightarrow$

$1 \propto \in k_B T$ and $l \propto \frac{1}{q^2}$

So, finding the dimension of the function

${\left[\frac{\varepsilon_0 k_B T}{q^2}\right]=\left[\frac{q^2 F r}{F r^2 q}\right]=\left[\frac{1}{r}\right]=\left[L^{-1}\right]}$

${\left[\frac{q^2}{\epsilon_0 k_B T}\right]=[L]}$

For option $A :\left[\sqrt{\frac{n q^2}{\varepsilon_0 k _{ B } T}}\right]=\left[\sqrt{ L ^{-3} L}\right]=\left[L^{-1}\right]$

For option $B :\left[\sqrt{\frac{\varepsilon_0 k _{ B } T}{ nq ^2}}\right]=\left[\sqrt{\frac{ L ^{-1}}{L^{-3}}}\right]=[ L ]$

For option C: $\left[\sqrt{\frac{ q ^2}{\varepsilon_0 K_{ B } T n^2}}\right]=\left[\sqrt{\frac{ L }{ L ^{-2}}}\right]=\left[ L ^{3 / 2}\right]$

For option D: $\left[\sqrt{\frac{q^2}{\varepsilon_0 K_{ B } n ^2}}\right]=\left[\sqrt{\frac{L}{L^{-1}}}\right]=[L]$

Hence, answer $B$ and $D$ matches correctly.