Trying to be certain

The probability of not getting a 6 after $n$ throws is $q = \left(\dfrac 56\right)^n$ . Therefore, the probability of getting at least a 6 after $n$ throws is $p = 1 - q = 1 - \left(\dfrac 56\right)^n$ . Then we have:

$\begin{aligned} p & \ge 0.\underbrace{999 \cdots 999}_{\text{\# of 9s }=26} \\ 1 - \left(\dfrac 56\right)^n & \ge 1 - 10^{-26} \\ \implies \left(\dfrac 56\right)^n & \le 10^{-26} & \small \color{#3D99F6} \text{Taking }\log_{10} \text{ both sides} \\ n \left(\log_{10} 5 - \log_{10} 6\right) & \le - 26 \end{aligned}$

$\begin{aligned} \implies n & \ge \frac {26}{\log_{10} 6 - \log_{10} 5} \approx 328.361 \\ n & = \boxed{329} \end{aligned}$