Explanation

(a) $R = R_{0}e^{- \lambda t}$
$\Rightarrow \ \ \ \ \ \left( \frac{1}{3} \right) = e^{- \lambda\ \times \ 3} = e^{- 3\lambda}\ \ \ \ \ \ \ \ \ldots(i)$
Again $\frac{R'}{R_{0}} = e^{- \lambda\ \times \ 9} = e^{- 9\lambda} = {(e^{- 3\lambda})}^{3}$
=$\left( \frac{1}{3} \right)^{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack From\ Eq.(i)\rbrack$
$\ = \frac{1}{27}$
$\Rightarrow$ $R’\ =$ $\frac{R_{0}}{27}$
Hence, in 9 days activity will become $\left( \frac{1}{27} \right)$ of the original value.