A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to:

Given retardation $a=-kx$.We know that acceleration $a=v\frac{dv}{dx}$.So,
$$v\frac{dv}{dx}=-kx$$
Integrating both sides with respect to $x$:
$${\int }_{v_i}^{v_f}vdv={\int }_0^x-kxdx$$
$${\left[\frac{v^2}{2}\right]}_{v_i}^{v_f}=-k{\left[\frac{x^2}{2}\right]}_0^x$$
$$\frac{1}{2}(v_f^2-v_i^2)=-\frac{1}{2}k{x}^2$$
Multiplying by mass $m$:
$$\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=-\frac{1}{2}mk{x}^2$$
Loss in kinetic energy $\Delta K=K_i-K_f=\frac{1}{2}mk{x}^2$.Therefore, the loss of kinetic energy is proportional to $x^2$.