A chemical reaction has a rate constant of 6.93 × 10^ -3 ~s^ -1 ....

The unit of the rate constant is $\mathrm{s}^{-1}$, indicating a first-order reaction.

The half-life of the reaction is given by:

$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{6.93 \times 10^{-3}} = 100 \mathrm{~s}$

For $93.75\%$ completion, the percentage of reactant remaining is:

$100\% - 93.75\% = 6.25\%$

The fraction of reactant remaining is:

$\dfrac{6.25}{100} = \dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$

This indicates that $4$ half-lives have passed.

Total time required $= 4 \times t_{1/2} = 4 \times 100 \mathrm{~s} = 400 \mathrm{~s}$

Answer: $400 \mathrm{~s}$