Correct Answer - B
Solving this question relative to the frame (car) of reference. For maximum velocity (relative to frame), the block must be in equilibrium position.
Let `x_0` be the equilibrium elongation in spring, then
`ma=kx_0`
From work-energy theorem,
`(mv^2)/(2)-0=-(kx_0^2)/(2)+ma x_0`
Solving the above equation, we get `v=asqrt(m/l)`
This question is an application of using the work-energy theorem in non-inertial frame of reference.