A constant force produces maximum velocity v on the block connected to the spring of force constant k as shown in the figure. When the force constant of spring becomes 4k,the maximum velocity of the block is (block is at rest when spring is relaxed): 

By work energy theorem;

$\begin{array}{l}F{x}_1-\frac{1}{2}k{x}_1^2=\frac{1}{2}m{v}^2 .......\left(1\right)\\ \mathrm{and} F{x}_2-\frac{1}{2}{k}^{\mathrm{\prime }}{x}_2^2=\frac{1}{2}m{v}^{\mathrm{\prime }2} .........\left(2\right)\end{array}$

where x₁,x₂ are initial and final extensions and v,t' are initial
and final velocities.
In both cases, force applied is same, and velocity becomes maximum when F=kx. (after which the mass will decelerate)

$\begin{array}{l}\therefore F=k{x}_1=\left(4k\right)x_2\\ \Rightarrow x_2=\frac{x_1}{4}\end{array}$

Substituting in (2):

$\begin{array}{l}F\left(\frac{x_1}{4}\right)-\frac{1}{2}\left(4K\right){\left(\frac{x_1}{4}\right)}^2=\frac{1}{2}m{v}^{\mathrm{\prime }2}\\ \Rightarrow \frac{1}{4}\left[F{x}_1-\frac{1}{2}k{x}_1^2\right]=\frac{1}{2}m{v}^{\mathrm{\prime }2} .......\left(3\right)\end{array}$

Dividing (3)/(1), we get:

$\frac{1}{4}=\frac{v^{\mathrm{\prime }2}}{v^2}\Rightarrow v^{\mathrm{\prime }}=\frac{v}{2}$

Hence, (c).