When dipped in water

$W_{\text {app }}=W_{\text {air }}-F_B$

$\Rightarrow 9 \,gm \times g=10 \,gm \times g-F_B$

Where,

$\left\{\begin{array}{l}V_c=\text { volume of cavity } \\ V_g=\text { volume of gold } \\ W_{\text {app }}=9 gm \\ W_{\text {air }}=10 gm \\ F_B=\text { force of buoyancy } \\ \rho_w=\text { density of water }=1 \\ \rho_g=\text { density of gold }=19.3\end{array}\right.$

$\Rightarrow 1 \times g=F_B$

Now (total volume displaced) $\times \rho_w \times g=1 \times g$

$\left(V_c+V_g\right) \times 1=1$

$V_c=1-\frac{\text { Mass of gold in air }}{\rho_g}=1-\frac{10}{19.3}=0.482 \,cc$