For a first-order reaction, the concentration of the reactant remaining after $n$ half-lives is given by the relation:

$[R] = \frac{[R]_0}{2^n}$

Given: Initial concentration, $[R]_0 = 1.28 \text{ mg L}^{-1}$ Final concentration, $[R] = 0.04 \text{ mg L}^{-1}$

Substituting these values into the equation: $0.04 = \frac{1.28}{2^n}$ $2^n = \frac{1.28}{0.04} = 32$ $2^n = 2^5$ Therefore, the number of half-lives, $n = 5$.

The total time required ($t$) is the product of the number of half-lives and the half-life period ($t_{1/2}$): $t = n \times t_{1/2}$ $t = 5 \times 138 \text{ s} = 690 \text{ s}$