Answer
See explanation.
Work Step by Step
$$\lambda=\frac{c}{f}=2.5 m$$
a. Use $\phi=(\frac{2\pi}{\lambda})(r_2-r_1)$ to relate the phase to the wavelength and the path difference.
$$\phi =(\frac{2\pi}{2.5m})(1.8m)=4.52 rad$$
b. Use $I=I_0cos^2(\phi/2)$
$$I=I_0cos^2(\frac{4.52rad}{2})=0.404I_0$$
