Answer
$K_C \lt K_A=K_D\lt K_B$
Work Step by Step
We know that
Kinetic energy for jogger A
$K.E_A=\frac{1}{2}mv^2=K.E$
Kinetic energy for jogger B
$K.E_B=\frac{1}{2}(\frac{m}{2})(3v)^2$
$K.E_B=\frac{9}{4}mv^2$
$K.E_B=\frac{9}{2}(\frac{1}{2}mv^2)=\frac{9}{2}K.E=4.5K.E$
Kinetic energy for jogger C
$K.E_C=\frac{1}{2}(3m)(\frac{v}{2})^2$
$K.E_C=\frac{3}{4}(\frac{1}{2}mv^2)$
$K.E_C=0.75K.E$
Kinetic energy for jogger D
$K.E_D=\frac{1}{2}(4m)(\frac{v}{2})^2$
$K.E_D=\frac{1}{2}mv^2=K.E$
Now we can rank the joggers in order of increasing kinetic energy as follows:
$K_C \lt K_A=K_D\lt K_B$
