To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as

Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as

where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then

Resolving for F,

PART B) We need here to apply the shear strength equation, then



In such a way that the material is more resistant to tensile strength than shear force.
Density: g/mL, kg/cubic meter
Volume: L, teaspoon
Mass: g, MeV/sq. C
The elastic potential energy of a spring is given by

where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:

In our problem, initially the spring is uncompressed, so

. Therefore, the work done by the spring when it is compressed until

is

And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.