Answer
$PE_e=4.4J$
Work Step by Step
Since the block hangs from the vertical spring, its weight pulls it down and equals the restoring force, so $$W=F_x=kx$$ $$x=\frac{W}{k}$$ which is how much the spring stretches.
Knowing $x$, we can get the elastic potential energy $$PE_e=\frac{1}{2}kx^2=\frac{1}{2}k\frac{W^2}{k^2}=\frac{W^2}{2k}$$
The weight of a 3.2kg block is $W=3.2\times9.8=31.36N$. The elastic potential energy $PE_e=1.8J$. So, $$\frac{31.36^2}{2k}=1.8$$ $$k=\frac{31.36^2}{3.6}=273.18N/m$$
The weight of a 5kg block is $W=5\times9.8=49N$. The elastic potential energy $PE_e$ is $$PE_e=\frac{W^2}{2k}=4.4J$$
