The potential energy of a long spring when stretched by 2cm is U ....

Potential energy stored in spring is given by,

$U = \dfrac{1}{2}K{x}^{2} .$

First case:

$U = \dfrac{1}{2}K{\left(2 \times {10}^{- 2}\right)}^{2} . . . Eq\left(1\right)$

Second case:

${U}^{'} = \dfrac{1}{2}K{\left(8 \times {10}^{- 2}\right)}^{2} . . . Eq\left(2\right)$

From both equations, we get

$\dfrac{U}{{U}^{'}} = \dfrac{1}{16} ⇒ {U}^{'} = 16U$