SrO is doped with 10^-4 mol % of Sr⁺⁴ The concentration of created cation vacancies is N_A × 10^x per mol. Value of x is ________.

When one atom of Sr⁴⁺ replace Sr²⁺, to maintain electrical neutrality, one Sr²⁺ ion removes. Thus, the no. of cation vacancy will be equal to added SrO₂ molecules.
In one mole of the oxide, no. of moles of SrO₂ added 

$=\frac{{10}^{-5}}{100}={10}^{-7}$

No. of cation vacancies = ${10}^{-7}\times 6.023\times {10}^{23}=6.023\times {10}^{16}$

$\therefore x = 16$