I think I would define a "twinkling of an eye" to be the amount of time it takes for reflected light to travel from the surface of the eyeball in question to an observer, which would of course radically depend on how far away the observer was.
But, regarding your computation, your issue is just keeping track of units, otherwise known as dimensional analysis. $299,792,458,000\frac{\text{mm}}{\text{sec}}$ is not the same thing as $299,792,458,000\text{ mm}$.
You have a length, $L=26.\bar{6}\text{ mm}$, and a velocity, $V=299,792,458,000\frac{\text{mm}}{\text{sec}}$. To get a time quantity, you would have to divide $\frac{L}{V}$, because this corresponds to computing
$$\frac{\text{length}}{\tfrac{\text{length}}{\text{time}}}=\frac{\text{length}\cdot\text{time}}{\text{length}}=\text{time}$$
It will have units of
$$\dfrac{\text{mm}}{\tfrac{\text{mm}}{\text{sec}}}=\frac{\text{mm}\cdot\text{sec}}{\text{mm}}=\text{sec}.$$
So, the result is
$$\frac{L}{V}\approx8.895\times 10^{-11}\text{sec}.$$
By Newton's second law (rate of change of momentum is equal to net force applied),
$$m \frac{dv}{dt} = mg - \alpha v$$; $\alpha $ is a constant, $\alpha v$ is the resisitive force.
Integrate w.r.t. time to get $v$ in terms of $t$.
$v|_{t = 0} = 0$ and $v|_{t \to \infty} = v_t$, terminal velocity is attained when $\frac{dv}{dt} = 0$, i.e. $v_t = \frac{mg}{\alpha}$
$m \int_{0}^{\frac{v_t}{2}} \frac{dv}{mg - \alpha v} = \int_{0}^{t_0}dt\quad$; $t_0$ is the required time.
Best Answer
0.59839860068 would be the result of the division, but it needs to be rounded to the least numerical accuracy of measurement values. 2 and $\pi$ are not measurement values, but 10.5 is.
10.5 has 3 "Significant figures".
So your answer also has to have 3 significant figures. The leading 0. do not count, so 0.598 would be the correct answer.