Total possibilities $=9 \times 10^{k-1},$ because most significant digit has $9$ options (excluding $0)$ and every other digit has $10$ options from $0$ to $9.$
Possibility of not containing any of the digits $0, 5, 9 = 7^{k},$ as now every digit has $7$ options.
Required probability $=\dfrac{7^{k}}{9 \times10^{k-1}} \approx (0.7)^k$
So C is the answer.