Current through the galvanometer,

$I_{G}=0.2 \% \text { of } I=\frac{0.2}{100} I=\frac{1}{500} I$

$\therefore$ Current through the shunt,

$I_{S}=I-I_{G}=I-\frac{1}{500} I=\frac{499}{500} I$

As shunt and galvanometer are in parallel

$\therefore \quad I_{G} G=I_{S} S$

$\left(\frac{1}{500} I\right) G=\left(\frac{499}{500}\right) S \text { or } S=\frac{G}{499}$

Resistance of the ammeter $R_{A}$ is

${\frac{1}{R_{A}}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{1}{G}=\frac{500}{G}}$

${R_{A}=\frac{1}{500} \,G}$