Given that

The mean length of the iron ring

l=50 cm=0.5 m

$l_{ag}=1mm=1\times {10}^{-3}m$

I=1 Amp.

Number of turns N=200

Relative permeability of iron ${\mu }_r=300$

For finding the flux density, the ampere turns for iron and air gape are to be determined. Leakage and fringing effects are neglected

At iron=$H\times l$

Where H is the magnetic field intensity

$H=\frac{B}{{\mu }_0{\mu }_r}=\frac{B}{4\pi \times {10}^{-7}\times 300}=2652.58BAT/m$

Where B is the flux density.

Hence AT for iron is $HlAT=2652.58B\times 0.5AT=26.29BAT$

At for air gap is $H_{airgap}\times l_{ag}AT=\frac{B}{4\pi \times {10}^{-7}}\times {10}^{-3}=795.77BAT$

Total ampere turns is $B(1326.29+795.77)AT=2122.06BAT$

Total ampere turns can be expressed as $N\times I=200\times 1AT200\times 1=2122.06B$

Hence flux density $B=\frac{200}{2122.06}Wb/m^2=0.094Wb/m^2$