Answer
We can rank the situations in order of the stopping potential, from largest to smallest:
$e \gt a = f \gt b = c \gt d$
Work Step by Step
Let $\lambda_0 = 50~nm$. We can write an expression for $K_{max}$, the maximum kinetic energy of an electron in the experiment:
$K_{max} = \frac{hc}{\lambda_0}-\phi = \frac{hc}{50~nm}-\phi = E_0-\phi$
We can write an expression for $K_{max}$ in each case:
(a) $K_{max} = \frac{hc}{200~nm}-\phi = \frac{1}{4}\times E_0-\phi$
(b) $K_{max} = \frac{hc}{250~nm}-\phi = \frac{1}{5}\times E_0-\phi$
(c) $K_{max} = \frac{hc}{250~nm}-\phi = \frac{1}{5}\times E_0-\phi$
(d) $K_{max} = \frac{hc}{300~nm}-\phi = \frac{1}{6}\times E_0-\phi$
(e) $K_{max} = \frac{hc}{100~nm}-\phi = \frac{1}{2}\times E_0-\phi$
(f) $K_{max} = \frac{hc}{200~nm}-\phi = \frac{1}{4}\times E_0-\phi$
Note that the stopping potential is proportional to the value of $K_{max}$ in each case.
We can rank the situations in order of the stopping potential, from largest to smallest:
$e \gt a = f \gt b = c \gt d$
