compressed by $(8.75-x)$

The compressed force is,

$F=k x$

$F=300 x=400(8.75-x)$

On solving the above equation,

$x=5 \,cm$

Spring $B$ is compressed by $(8.75-5)=3.75 cm$

Now,

$\frac{E_{A}}{E_{B}}=\frac{\frac{1}{2} k_{A} x_{A}^{2}}{\frac{1}{2} k_{B} x_{B}^{2}}$

$=\frac{\frac{1}{2} \times 300 \times 5^{2}}{\frac{1}{2} \times 400 \times 3.75^{2}}$

$=\frac{4}{3}$