Answer: 25.71: 1
Let the Frenkel defect be $n$ in the ionic crystal of $N$ ions within interstitial space, where $n = (NN_i)^{1/2}\;$$ e^{\large\frac{-E}{2kT}}$
Since for this crystal, $N_i = 2N \rightarrow n = (2N^2)^{1/2}\;$$ e^{\large\frac{-E}{2kT}} = \sqrt 2 N e^{\large\frac{-E}{2kT}}$
$\big (\large\frac{n}{N} $$\big )$$_\text{Frenkel} = \sqrt 2 \times e ^ {\large\frac{-2 \times 1.6 \times 10^{-19}} {2 \times 1.23 \times 10^{-23} \times 4000}}$$ = \sqrt 2 e^{-2.8985}$
It is given that the amount of energy needed to form Schottky defects ($4\;eV$) is twice that of what's needed for Frenkel defect ($2\; eV$).
We can use what we just calculated and write the ratio as follows:
$\big (\large\frac{n}{N} $$\big )$$_\text{Schottky} = \sqrt 2 \times e ^ {\large\frac{-4 \times 1.6 \times 10^{-19}} {2 \times 1.23 \times 10^{-23} \times 4000}}$$ = \sqrt 2 e^{-5.797}$
Therefore the ratio of mole fractios of Frenkel and Schottky's defects is: $\large\frac{\sqrt 2 e^{-2.8985}} { \sqrt 2 e^{-5.797}}$$ = 25.71$