Answer:- 

For a first-order reaction:

t = 2.303/k * log₁₀[A]₀/[A]_t

i. Time is taken for 99.9% completion:

Let the time taken for 99.9% completion of the reaction be t_99.9%.

Let initial concentration, [A]₀ = a

The final concentration, [A]_t = a - 99.9% of a

= a - (99.9/100 * a) = 0.001 * a

t_99.9% = 2.303/k * log₁₀[A]₀/[A]_t

= 2.303/k * log₁₀ a/(0.001 * a)

= 2.303/k * log₁₀ 1000 ... (1)

ii. Time is taken for 90% completion:

Let the time taken for 90% completion of the reaction be t_90%.

Let initial concentration, [A]₀ = a

Then, final concentration, [A]_t = a - 90% of a

= a - (90/100 * a) = 0.1 * a

t_90% = 2.303/k * log₁₀[A]₀/[A]_t = 2.303/k * log₁₀ a/(0.1 * a)

= 2.303/k * log₁₀ 10 ... (2)

Dividing (1) by (2), we get

(t_99.9%/t_90%) = (2.303/k * log₁₀ 1000)/(2.303/k * log₁₀ 10) = (log₁₀ 1000)/(log₁₀ 10) = 3/1

Therefore, (t_99.9%/t_90%) = 3

So, t_99.9% = 3 * t_90%

Therefore, for a first-order reaction, the time required for 99.9% completion is 3 times that required for 90% completion.