\(\Rightarrow {i}_{2}=\frac{5}{{R}_{{L}}}\)

Also voltage across \({R}=50-5=45\) \(volt\)

By \(v=i R \Rightarrow R=\frac{v}{i}=\frac{45}{i_{1}+i_{2}}\)

\({R}=\frac{45}{90 {m} {A}+\frac{5}{{R}_{{L}}}}\)

Current in zener diode is maximum when \({R}_{{L}} \rightarrow \infty\)

\(\left(i_{2} \rightarrow 0\right.\) and \(\left.i_{i}=i\right)\)

So \({R}=\frac{45}{90 {mA}}=500 \,\Omega\)