Answer
$(U_s)_d > (U_s)_c > (U_s)_b = (U_s)_a$
Work Step by Step
When a spring is compressed a distance of $x$, the elastic potential energy stored in the spring is $U_s = \frac{1}{2}kx^2$, where $k$ is the spring constant.
We can find the elastic potential energy stored in spring a.
$(U_s)_a = \frac{1}{2}kd^2$
We can find the elastic potential energy stored in spring b.
$(U_s)_b = \frac{1}{2}kd^2$
We can find the elastic potential energy stored in spring c.
$(U_s)_c = \frac{1}{2}(2k)d^2$
$(U_s)_c = kd^2$
We can find the elastic potential energy stored in spring d.
$(U_s)_d = \frac{1}{2}k(2d)^2$
$(U_s)_d = 2kd^2$
We can rank elastic potential energy in the springs in order from most to least:
$(U_s)_d > (U_s)_c > (U_s)_b = (U_s)_a$
