$R =0.005 \Omega, r _2=0.1 \ m$

$I = I _0 \cos (300 t )$

So we get $\omega=300$

The magnetic field of the tube, $B=\frac{\mu_0 NI }{ L }$

In the problem $N =1, \quad L=10 m, \quad I = I _0 \cos (300 t$ )

So, $B=\frac{\mu_0 I_0 \cos (300 t)}{10}$

$Q=\frac{\mu_0 I_0 \cos (300 t)}{10} \times \pi(0.1)^2$

$\text { Induced emf, } e=-\frac{d Q}{d t}=300 \mu_0 \pi I_0 \sin (300 t) \times 10^{-3}$

$e=0.3 \mu_0 \pi I_0 \sin (300 t)$

The current in the ring $i =\frac{ e }{ R }$

$i =\frac{0.3 \pi \mu_0 I _0 \sin (300 t)}{0.005}$

$\text { or } i =60 \mu_0 \pi I _0 \sin (300 t)$

$\text { Magnetic moment, } M = iA = i \pi r ^2$

$M =60 \mu_0 \pi I _0 \sin (300 t) \times \pi(0.1)^2$

$=0.6 \mu_0 \pi^2 I _0 \sin (300 t) A$

$=\left(0.6 \times \pi^2\right) \mu_0 I _0 \sin (300 t)$

$=6 \mu_0 I _0 \sin (300 t)$

So, the value of $N =6$