An electron accelerated through a potential difference V 1 has a de-Broglie wavelength of λ . When the…

Let initial wavelength be $\lambda$, then after $50\%$ increase wavelength will become $1.5\lambda$.

Now, $KE=\frac{P^2}{2m}=eV \& P=\frac{h}{\lambda }$

Therefore, $eV=\frac{{\left({\displaystyle \frac{h}{\lambda }}\right)}^2}{2m}$.

So we can write

$e{V}_1=\frac{{\left({\displaystyle \frac{h}{\lambda }}\right)}^2}{2m} ...Eq\left(1\right)$ and 

$e{V}_2=\frac{{\left({\displaystyle \frac{h}{1.5\lambda }}\right)}^2}{2m} ...Eq\left(2\right)$

From both equations, we get

$\frac{V_1}{V_2}={\left(1.5\right)}^2=\frac{9}{4}$