The number of real values of the parameter k for which ${({\log }_{16} \mathrm{x})}^2-{\log }_{16} \mathrm{x}+{\log }_{16} \mathrm{k}=0$ with real coefficients will have exactly one solution is

Let ${\log }_{16} \mathrm{x}=\mathrm{y}\Rightarrow {\mathrm{y}}^2-\mathrm{y}+{\log }_{16} \mathrm{k}=0$

This quadratic equation will have exactly one solution if its discriminant vanishes.

$\therefore {(-1)}^2-4.1.{\log }_{16} \mathrm{k}=0\Rightarrow 1={\log }_{16} {\mathrm{k}}^4$

$\Rightarrow {\mathrm{k}}^4=16\Rightarrow {\mathrm{k}}^2=4\Rightarrow \mathrm{k}=\pm 2.$

But ${\log }_{16} \mathrm{k}$ is not defined $\mathrm{k}<0, \therefore \mathrm{k}=2.$

$\therefore$ Number of real values of $\mathrm{k}=1.$