In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be 
 

For a converted ammeter, i_gG = i_sS
where G and S are the resistances of the galvanometer and shunt, and i_g and i_s are their respective currents.
Here, ${\mathrm{i}}_g=\frac{0.2\mathrm{l}}{100}{\mathrm{i}}_{\mathrm{s}}=\frac{99.8\mathrm{I}}{100}=0.2\mathrm{G}=99.8\mathrm{S}$
$S=\frac{1}{499}G$
as S and G are in parallel, resistance of the ammeter,
$R_A=\frac{GS}{G+S}=\frac{G(G/499)}{G+\frac{G}{499}}=\frac{G}{500}$