Do you understand that you're trying to do $\ce{CaCl2.6H2O -> CaCl2.2H2O + 4H2O}$? Just assume that in the beginning you had $\pu{100 gm}$ of the total mixture. So that makes the weight of $\ce{CaCl2.6H2O}$ present in the intial mixture to be:
$27\%$ of $\pu{100 gm = 27gm}$.
Next, calculate the number of moles of $\ce{CaCl2.6H2O}$ you had in the beginning by:
dividing the mass you've just obtained by the molecular mass of this compound.
Now according to the stoichiometry of the reaction I've given you, when $1$ mole of $\ce{CaCl2.6H2O}$ converts, you get $1$ mole of $\ce{CaCl2.2H2O}$. Can you calculate the mass of $\ce{CaCl2.2H2O}$ thus obtained? So now you've the mass of $\ce{CaCl2.6H2O}$ and $\ce{CaCl2.2H2O}$.
To calculate the percentage of $\ce{CaCl2}$ in $\ce{CaCl2.2H2O}$ all you've got to do is:
$\displaystyle \frac{\text{Mass of } \ce{CaCl2}}{\text{Mass of } \ce{CaCl2.2H2O}} \times 100\%$
You can carry out this step for $\ce{CaCl2.6H2O}$ on your own and then finally calculate the percentage change.