Answer:
Approximately [tex]13\; {\rm N \cdot m^{-1}}[/tex] (assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].)
Explanation:
Let [tex]F_{\text{s}}[/tex] denote the force that this spring exerts on the object. Let [tex]x[/tex] denote the displacement of this spring from the equilibrium position.
By Hooke's Law, the spring constant [tex]k[/tex] of this spring would ensure that [tex]F_\text{s} = -k\, x[/tex].
Note that the mass of the object attached to this spring is [tex]m = 540\; {\rm g} = 0.540\; {\rm kg}[/tex]. Thus, the weight of this object would be [tex]m\, g = 0.540\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \approx 5.230\; {\rm N}[/tex].
Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, [tex]F_{\text{s}} = 5.230\; {\rm N}[/tex].
The spring in this question was stretched downward from its equilibrium by:
[tex]\begin{aligned} x &= (70\; {\rm cm} - 110\; {\rm cm}) \\ &= (-40)\; {\rm cm} \\ &= (-0.40) \; {\rm m}\end{aligned}[/tex].
(Note that [tex]x[/tex] is negative since this displacement points downwards.)
Rearrange Hooke's Law to find [tex]k[/tex] in terms of [tex]F_{\text{s}}[/tex] and [tex]x[/tex]:
[tex]\begin{aligned} k &= \frac{F_{\text{s}}}{-x} \\ &\approx \frac{5.230\; {\rm N}}{-(-0.40)\; {\rm m}} \\ &\approx 13\; {\rm N \cdot m^{-1}}\end{aligned}[/tex].
