Weight of anhydrous $BaCl_2 = 52\,g$

Loss in mass $= 9\,g$

Assuming $BaCl_2.xH_2O$ as hydrate mass of $H_2O = 9\,g$

Moles of ${{H}_{2}}O=\frac{9}{18}=0.5$

Grass molecular let of $BaCl_2 = 208$

$\%$ of $H_2O$ in this hydrated $BaC{{l}_{2}}=\frac{9}{61}\times 100$

$=14.75\%$

$=\frac{18x}{208+18x}\times 100$ on solving $x = 2$

This percentage is present in $BaCl_2.2H_2O$