Answer
$\Phi_B=0.022Wb$
Work Step by Step
The magnetic flux is equal to $$\Phi_B=BAcos\theta$$ Since theta is the angle below the horizontal and since it is supposed to be the angle from the normal, subtract the old angle from 90 degrees to get $\theta=90^{\circ}-72^{\circ}=18^{\circ}$ Substituting this value along with $A=lw=(22m)(18m)=396m^2$ and $B=5.9\times 10^{-5} T$ yields a flux of $$\Phi_B=(5.9\times 10^{-5}T)(396m^2)cos(18^{\circ})=0.022Wb$$
