A 8V Zener diode along with a series resistance R is connected...

Let ${I}_{1}$ be the current through Zener diode and ${I}_{2}$ through load resistor.

As maximum current flows through Zener diode, hence ${I}_{2} = 0$.

∴Maximum current, $I = {I}_{1} + {I}_{2} = 25mA + 0 = 25mA$

Here, ${V}_{o}$ is voltage drop across Zener $= 8V$.
Since $I$ is maximum, hence minimum value of $R$

R = \dfrac{{V}_{i} -}{{V}_{o}} = \dfrac{20 - 8}{25 × {10}^{- 3}} = 480Ω