Let when the combination of springs is compressed by force \(\mathrm{F}\). Spring \(A\) is compressed by \(x\). Therefore compression in spring \(\mathrm{B}\)

\(x_{B}=(8.75-x) c m\)

\(\mathrm{F}=300 \times x=400(8.75-x)\)

Solving we get, \(x=5 \mathrm{cm}\)

\(x_{B}=8.75-5=3.75 \mathrm{cm}\)

\(\frac{E_{A}}{E_{B}}=\frac{\frac{1}{2} k_{A}\left(x_{A}\right)^{2}}{\frac{1}{2} k_{B}\left(x_{B}\right)^{2}}=\frac{300 \times(5)^{2}}{400 \times(3.75)^{2}}=\frac{4}{3}\)