Specific gravity of gold is 19.3 and that of copper is 8.9.

Data: m$_{air}$=36 gm

m$_{water}$=34 gm

specific gravity of gold=S$_{gold}$=19.3

specific gravity of copper=S$_{copper}$=8.9

∴ Weight of ornament in air is=W$_{air}$=$36 \times 10^{-3} \times 9.81$=0.35316 N

∴ Weight of ornament in water is=W$_{water}$=$34 \times 10^{-3} \times 9.81$=0.33354 N

Buoyant force=$F_{B}=W_{air}-W_{water}$=0.01962 N

Weight of the fluid displaced $(F_{B})=W_{water} \times $volume of the ornament

0.01962=9810$\times $volume of the ornament

∴ Volume of the ornament=$2 \times 10^{-6} m^{3}$

But, the ornament contains gold and copper

∴ Volume of the ornament (V) = $V_{gold}+V_{copper}$ ---------------(1)

$V_{gold}=\cfrac{\text{Weight of gold}}{W_{gold}}$

$V_{gold}=\cfrac{\text{Weight of gold}}{19.3 \times 9810}$

Similarly,

$V_{copper}=\cfrac{\text{Weight of copper}}{W_{copper}}$

$V_{copper}=\cfrac{\text{Weight of copper}}{8.9 \times 9810}$

Equation (1) becomes,

$2 \times 10^{-6}$= $V_{gold}+V_{copper}$

$2 \times 10^{-6}$=$\cfrac{\text{Weight of gold}}{19.3 \times 9810} + \cfrac{\text{Weight of copper}}{8.9 \times 9810}$ ---------------(2)

Also, weight of the ornament (V) = $W_{gold}+W_{copper}$=0.35316 ---------------(3)

Solving equation (2) and (3) we get,

$W_{gold}$=0.33133 N and $W_{copper}$=0.0218 N