To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.
The extension of the spring due to the weight of the object on Earth is 0.3m, then
[tex]F_k = F_{W,E}[/tex]
[tex]kx_1 = mg[/tex]
The extension of the spring due to the weight of the object on Moon is a value of [tex]x_2[/tex], then
[tex]kx_2 = mg_m[/tex]
Recall that gravity on the moon is a sixth of Earth's gravity.
[tex]kx_2 = m\frac{g}{6}[/tex]
[tex]kx_2 = \frac{1}{6} mg[/tex]
[tex]kx_2 = \frac{1}{6} kx_1[/tex]
[tex]x_2 = \frac{1}{6} x_1[/tex]
We have that the displacement at the earth was [tex]x_1 = 0.3m[/tex], then
[tex]x_2 = \frac{1}{6} 0.3[/tex]
[tex]x_2 = 0.05m[/tex]
Therefore the displacement of the mass on the spring on Moon is 0.05m
