The half life period of a first order chemical reaction is $$6.93$$ minutes. The time required for the completion of $$99\%$$ of the chemical reaction will be ($$\log 2 = 0.301$$) :

For a first-order reaction the rate constant $$k$$ is related to the half-life $$t_{1/2}$$ by

$$k = \frac{0.693}{t_{1/2}} \qquad -(1)$$

Given $$t_{1/2} = 6.93\text{ min}$$, substitute in (1):
$$k = \frac{0.693}{6.93} = 0.100\;\text{min}^{-1}$$

Let the initial concentration be $$[A]_0$$ and the concentration after time $$t$$ be $$[A]$$. For a first-order reaction

$$\ln\!\left(\frac{[A]_0}{[A]}\right)=kt \qquad -(2)$$

Completion of $$99\%$$ of the reaction means $$1\%$$ is left, i.e. $$[A] = 0.01[A]_0$$.
Hence $$\dfrac{[A]_0}{[A]} = \dfrac{[A]_0}{0.01[A]_0}=100$$.

Insert this value in (2):
$$\ln(100)=kt$$

Using $$\ln(100)=2\ln(10)=2\times2.303=4.606$$, we get

$$t = \frac{4.606}{k} = \frac{4.606}{0.100}=46.06\text{ min}$$

Therefore, the time required for $$99\%$$ completion is

Option C which is: $$46.06$$ minutes