Answer
$11.7s$
Work Step by Step
The axis of rotation is at the center of the reel. Tension $T$ produces a torque with a lever arm being the radius of the rotation $r=0.16m$. This torque is in opposite direction with the torque produce by friction $f$ $$\sum\tau=\tau_T-\tau_f=Tr-\tau_f$$ $$\sum\tau=25\times0.16-3.4=0.6N.m$$
The angular acceleration of the reel is $$\alpha=\frac{\sum\tau}{I_{reel}}=\frac{0.6}{0.44}=1.36rad/s^2$$
The hose unwinds without slipping, so the rotation is a rolling motion, meaning the linear acceleration of the reel is $$a=a_T=r\alpha=0.22m/s^2$$
We have the reel's initial speed $v_0=0$, and the length of the hose is $s=15m$. The time it takes to unwind the hose is $$s=v_0t+\frac{1}{2}at^2=\frac{1}{2}at^2$$ $$t=\sqrt{\frac{2s}{a}}=11.7s$$
