Considering an element of tube of length dx having mass dm at a distance of x from right end as shown in figure below.

Now, force acting of that element is F=∫(dm)ω2x
Integrating the above relation for whole length of tube, we have
F=∫0L(Lmdx)ω2x
=Lmω22L2 =2mω2L
⇒ω=mL2F =0.25×0.52F
=16F=4Frads−1
Hence, the value of x=4.