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

As shown in figure a shunt $S$ is connected in parallel with the galvanometer to convert it into the ammeter, let $G$ is the resistant of the galvanometer. Where $I$ is the current in the circuit as shown, according to question $0·2\%$ current passing through the galvanometer, i.e., 

$I_G = 0·2\% of I = \frac{0·2}{100}I= \frac{I}{500}$;

so, the current through the shunt is $$\begin{gathered} I=I_G + I_S \Rightarrow I_S = I-I_G \\ \Rightarrow I_S = I-\frac{1}{500} =\frac{499}{500}I \end{gathered}$$ 
As $\left(G\right)$ and the shunt $\left(S\right)$ are in parallel combination ,

$V_G = V_S$

$i_g{R}_g=i_s{R}_s\Rightarrow \left(\frac{i}{500}\right)\left(G\right)=\left(\frac{499}{500}i\right)\left(S\right)$
$\Rightarrow S=\frac{G}{499}$
Equivalent resistance of the ammeter
$\frac{1}{R_{\mathrm{e}\mathrm{q}}}=\frac{1}{G}+\frac{1}{\frac{G}{499}}\Rightarrow R_{\mathrm{e}\mathrm{q}}=\frac{G}{500}$