Answer
System B's spring constant is four times that of system A.
Work Step by Step
We know the total energy of a spring mass system $(E)=\frac{1}{2}KA^{2}$ , Where. K - spring constant, a - amplitude of the oscillation.
For the system A, let's assume $K=K_{1}, A=A$ & for the system B, $K=K_{2},\space A=A/2$ & apply above equation to both systems.
$E_{A}=\frac{1}{2}K_{1}A^{2}-(1)$
$E_{B}=\frac{1}{2}K_{2}(\frac{A}{2})^{2}-(2)$
(1)=(2) =>
$\frac{1}{2}K_{1}A^{2}=\frac{1}{2}K_{2}(\frac{A}{2})^{2}$
$K_{1}=\frac{K_{2}}{4}=>4K_{1}=K_{2}$
System B's spring constant is four times that of system A