Time required for completion of 99.9% of first order reaction is _____ times of half life ($$t_{1/2}$$) of the reaction

For a first-order reaction: $$t = \frac{2.303}{k}\log\frac{1}{1-x}$$

For 99.9% completion ($$x = 0.999$$):

$$t_{99.9\%} = \frac{2.303}{k}\log\frac{1}{0.001} = \frac{2.303}{k} \times 3 = \frac{6.909}{k}$$

$$t_{1/2} = \frac{0.693}{k}$$

$$\frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909}{0.693} \approx 9.97 \approx 10$$

The answer is $$\boxed{10}$$.