No calculus but calc

$x = y^{\frac{1}{y}}$

taking log on both sides:

$ln x = {\dfrac{1}{y}} ln y$

differentiating with respect to $y$ .

$\dfrac{1}{x} \dfrac{dx}{dy} = (\dfrac{1}{y} . \dfrac{1}{y} - \dfrac{1}{y^{2}} .ln y)$

Put $\dfrac{dx}{dy} = 0$

$0 = (\dfrac{1}{y^2} - \dfrac{ln y}{y^{2}}$

from here $y= e$ , and will be maximum at $e$

Now, $A = e^{\frac{1}{e}} = 1.44466$

$\sqrt{A} = \boxed{1.2019}$