For a function to be Rienmann integral, its upper integral and lower integral need to be the same. I don't understand the function they have given here where $x=r/2^n$ thus don't know how to apply to values of the bounds to the functions. Also how to work out the upper and lower sum from this?
On
For any fixed $n$ the function is zero on the whole interval up to $0,1/2^n,2/2^n,\dots, (2^n-1)/2^n,1$, where its value is $1$.
For example is $n=3$, then your function $f_3$ is $1$ at $0,1/8,2/8,3/8,4/8,5/8,6/8,7/8,8/8$ and zero everywhere else.
$f$ looks basically like $f_n$ but has more peaks: it is $1$ where $x=r/2^n$, $r,n\in\mathbb{N}$. So it is $1$ at $0,3/4,7/8,10/1224\ $ but not at $5/7,9/10,3/6\ $ or any irrational number.
Hint: for each fixed $n$ there is only a finite number of points of $x\in[0,1]$ s.t. $x = r/2^n$, $r\in\Bbb Z$. But the set of points $r/2^n\in[0,1]$ for some $r,n\in\Bbb Z$ is dense in $[0,1]$ (and in any subinterval).