Solution

In the given wave equation the particle's displacement amplitude $$=ae^{-\gamma x}$$
Let two points $$x_1$$ and $$x_2$$, between which the displacement amplitude differ by $$\eta =1\%$$
So, $$ae^{-\gamma x_1}-ae^{e^{-\gamma x_2}}=\eta ae^{-\gamma x_1}$$
or $$e^{-\gamma x_1}( 1-\eta )=e^{-\gamma x_2}$$
or $$\ln (1-\eta )-\gamma x_1 =-\gamma x_2$$
or, $$x_2 -x_1 =-\dfrac{\ln (1-\eta)}{\gamma}$$
So path difference $$=-\dfrac{\ln (1-\eta )}{\gamma}$$
and phase difference $$=\dfrac {2\pi}{\lambda}\times $$ path difference
$$=-\dfrac{2\pi}{\lambda}\dfrac{\ln (1-\eta )}{\gamma}=\dfrac{2\pi \eta }{\lambda. \gamma}=0.3\ rad$$