Answer
$\Delta x'=\frac{1}{2}\Delta x$
Work Step by Step
According to conservation of energy, upon the collision, the kinetic energy must be converted completely to elastic potential energy. This means that $$\frac{1}{2}k\Delta x^2=\frac{1}{2}mv^2$$ Solving for $\Delta x$ yields $$\Delta x=\sqrt{\frac{mv^2}{k}}=v\sqrt{\frac{m}{k}}$$ If the spring constant is increased by a factor of four, the new compression distance is $$\Delta x'=v\sqrt{\frac{m}{4k}}=\frac{v}{2} \sqrt{\frac{m}{k}}=\frac{1}{2}\Delta x$$
