The de-Broglie wavelength of a particle having kinetic energy E is...

Recall the formula of de-Broglie wavelength, $\lambda = \dfrac{h}{p}$, here $p$ is the linear momentum of the particle, now use the relation between linear momentum and kinetic energy, $p = \sqrt{2Em} ⇒ E ∝ \dfrac{1}{{\lambda}^{2}}$$⇒ E = \dfrac{k}{{\lambda}^{2}}$

$\text{and}{E}^{'} = \dfrac{16k}{9{\lambda}^{2}} ⇒ \DeltaE = \dfrac{7E}{9}$.