Answer:

[tex] \bold{Option \: (b)} [/tex]

Explanation:

The formula for the potential energy [tex] \sf (U) [/tex] stored in a spring is given by [tex] \sf U = \dfrac{1}{2}kx^2 [/tex].

[tex] \sf k = 2 \: N m^{-1} [/tex]

[tex] \sf x = 5×10^{-2} \: m [/tex]

[tex] \sf U= \dfrac{1}{2} × 2×5×5 × 10^{-2} × 10^{-2} [/tex]

[tex] \sf U = 25× 10^{-4} J = 0.0025 J = \bold{option \: (b)} [/tex]