The activity of a radioactive material is 6 . 4 × 10 ^ - 4 curie....

Given here, ${A}_{0} = 6 . 4 \times {10}^{- 4}$ Curie and $A = 5 \times {10}^{- 6}$ Curie

Half life, $T\dfrac{1}{2} = 5$ days

The activity of a radioactive sample at time $t$ is given by,

$A = {A}_{0}\left(\dfrac{1}{2}\right)\dfrac{t}{{T}_{\dfrac{1}{2}}}$

$⇒ \dfrac{5 \times {10}^{- 6}}{6 . 4 \times {10}^{- 4}} = \left(\dfrac{1}{2}\right)\dfrac{t}{5}$

$⇒ 7 = \dfrac{t}{5} ⇒ t = 35days$