If $\frac{{\log }_{2}a}{2}=\frac{{\log }_{3}b}{3}=\frac{{\log }_{4}c}{4}$ and $a^{1/2}{b}^{1/3}{c}^{1/4}=24$ then

$\begin{array}{l}\frac{{\log }_{2}a}{2}=\frac{{\log }_{3}b}{3}=\frac{{\log }_{4}c}{4}=\lambda \text{ (say) }\\ \Rightarrow a=2^{2\lambda },b=3^{3\lambda },c=4^{4\lambda }\end{array}$

Now,

$\begin{array}{l}{a}^{1/2}{b}^{1/3}{c}^{1/4}=24\\ \Rightarrow 2^{\lambda }\times 3^{\lambda }\times 4^{\lambda }=24\Rightarrow 2^{\lambda }\times 3^{\lambda }\times 4^{\lambda }=2\times 3\times 4\Rightarrow \lambda =1\end{array}$

Hence, $a=2^2=4,b=3^3=27$ and $c=4^4=256$