Number of turns on a ferromagnetic core, $N=3500$

Relative permeability of the core material, $\mu_{r}=800$

Magnetizing current, $I=1.2\, A$

The magnetic field is given by the relation:

$B=\frac{\mu_{r} \mu_{0} I N}{2 \pi r}$

Where, $\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \operatorname{Tm}\, A ^{-1}$

$B=\frac{800 \times 4 \pi \times 10^{-7} \times 1.2 \times 3500}{2 \pi \times 0.15}=4.48 \,T$

Therefore, the magnetic field in the core is $4.48 \,T$