This question picking numbers does not work very well

Better to use a general approach

We do know that $x=y^3$

Substitute in the two quantities

we do have

QA $y^{{y{^3}}$

QB $y^{y+y+y}$

We do have the same base so is the same to write

$y^3>3y$

$y^3-3y>0$

$y(y^2-3)>0$

Now

$y>0$ which is false because we do know that $y>1$

So we have to consider only $y^2>3$

Take the square root we do have that

$y>1.7$

We do know that $y>1$

However, we do have also $y>1.7$

Therefore y can be 1.1 or 2. Below 1.7 or greater than 1.7

The answer is D