For a first order reaction, the time required for completion of $$90\%$$ reaction is '$$x$$' times the half life of the reaction. The value of '$$x$$' is
(Given: $$\ln 10 = 2.303$$ and $$\log 2 = 0.3010$$)

For a first order reaction, the integrated rate law is: $$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$$.

When 90% of the reaction is complete, 10% of the reactant remains, so $$[A] = 0.1[A]_0$$. Accordingly, $$t_{90\%} = \frac{2.303}{k} \log \frac{[A]_0}{0.1[A]_0} = \frac{2.303}{k} \log 10 = \frac{2.303}{k}$$.

For a first order reaction, the half-life is given by $$t_{1/2} = \frac{0.693}{k} = \frac{2.303 \times \log 2}{k} = \frac{2.303 \times 0.3010}{k} = \frac{0.6932}{k}$$.

Therefore, the ratio is $$x = \frac{t_{90\%}}{t_{1/2}} = \frac{2.303/k}{0.693/k} = \frac{2.303}{0.693}$$ and hence $$x = \frac{2.303}{0.693} = 3.32$$.

The correct answer is Option C: 3.32.