Answer
The amount of work that must be done to stretch the spring is $1.60~J$
Work Step by Step
We can find the elastic potential energy stored in the spring when it is pulled back:
$U_s = \frac{1}{2}kx^2$
$U_s = \frac{1}{2}(20.0~N/m)(0.40~m)^2$
$U_s = 1.60~J$
The amount of work that must be done to stretch the spring is $1.60~J$
