$k=2 \times 10^{6} N / m$

$E_{T}=160 J$

The total mechanical energy,

$E_{T}=(K . E)_{\max }+(P . E)_{\min }=160 J \ldots \ldots(1)$

we know that, maximum kinetic energy,

$(K . E)_{\max }=\frac{1}{2} k A^{2} \frac{1}{2} \times 2 \times 10^{6} \times 10^{-6}$

$(K . E)_{\max }=100 J$

from equation $(1),$

$(P . E)_{\min }=160-100=60 J$

The minimum kinetic energy is zero so, the maximum potential energy is $160 \mathrm{J}$