Let G be the resistance of the galvanometer and Ig be the current required for full scale deflection.
When the galvanometer is shunted with S=2Ω, it acts as an ammeter of range I=500 mA=0.5 A. The potential difference across the galvanometer and the shunt is equal:
IgG=(I−Ig)S
IgG=(0.5−Ig)×2
IgG+2Ig=1
Ig=G+21
When a resistance R=470Ω is connected in series, it acts as a voltmeter of range V=10 V. The total voltage is the sum of the voltage drops:
V=Ig(G+R)
10=Ig(G+470)
Ig=G+47010
Equating the two expressions for Ig:
G+21=G+47010
G+470=10(G+2)
G+470=10G+20
9G=450
G=50Ω
Answer: 50