Answer
$\rm 4.0\;cm,\;34.7\;cm,\;65.3\;cm$
Work Step by Step
First of all, we need to find the initial full length of the tube,
$$L_i=2(80)+\frac{1}{2}(2\pi r)$$
where $\frac{1}{2}(2\pi r)$ is the circumference of a semicircle which is the length of the bent part of the tube.
We can see that $r=\frac{10}{2}=5$ cm since the distance between the two ends is 10 cm which is, at the same time, the diameter of the semicircle part of the tube.
Thus,
$$L_i=160+ 5\pi =\bf 175.7\;\rm cm$$
Now the final length is then given by
$$L = (\bf 2\it s+\bf 1.757 )\;\rm m\tag 1$$
To have a loud sound, we have to create a standing wave inside this open-open tube.
The frequency in this case is given by
$$f_m=\dfrac{mv}{2L}\tag {$m=1,2,3,...$}$$
Plugging from (1);
$$f_m=\dfrac{mv}{2( 2 s+ 1.757 )} $$
Solving for $s$;
$$2 s+ 1.757 =\dfrac{mv}{2 f_m} $$
$$2 s =\dfrac{mv}{2 f_m}-1.757 $$
$$ s =\dfrac{mv}{4 f_m}-\dfrac{1.757 }{2}$$
Plugging the known;
$$ s =\dfrac{343m}{4 \cdot 280}-\dfrac{1.757 }{2}$$
At $m=1$,
$$ s =\dfrac{343(1)}{4 \cdot 280}-\dfrac{1.757 }{2}=\bf -0.57\;\rm m
$$
Dismissed since the extension must be greater than zero and less than 80 cm.
At $m=2$,
$$ s =\dfrac{343(2)}{4 \cdot 280}-\dfrac{1.757 }{2}=\bf -0.266\;\rm m
$$
Dismissed since the extension must be greater than zero and less than 80 cm.
At $m=3$,
$$ s =\dfrac{343(3)}{4 \cdot 280}-\dfrac{1.757 }{2}=\bf 0.04025\;\rm m
$$
$$s=\color{red}{\bf 4.0}\;\rm cm$$
At $m=4$,
$$ s =\dfrac{343(4)}{4 \cdot 280}-\dfrac{1.757 }{2}=\bf 0.3465\;\rm m
$$
$$s=\color{red}{\bf 34.65}\;\rm cm$$
At $m=5$,
$$ s =\dfrac{343(5)}{4 \cdot 280}-\dfrac{1.757 }{2}=\bf 0.653\;\rm m
$$
$$s=\color{red}{\bf 65.3}\;\rm cm$$
At $m=6$,
$$ s =\dfrac{343(6)}{4 \cdot 280}-\dfrac{1.757 }{2}=\bf 0.959\;\rm m
$$
Dismissed since the extension must be less than 80 cm.
Therefore, we got only 3 slides that create a standing wave resonance which are the red results above.
