Let $I_1$ be the current through Zener diode and $I_2$ through load resistor.

​​​​​​​

As maximum current flows through Zener diode, hence $I_2=0$.

$\therefore$Maximum current, $I=I_1+I_2=25 \mathrm{mA}+0=25 \mathrm{mA}$

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

$R=\frac{V_i-}{V_o}=\frac{20-8}{25\times {10}^{-3}}=480 \mathrm{\Omega }$