c
Here, resistance of the galvanometer \(=G\)

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

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

\({R_{A}=\frac{1}{500} \,G}\)