A first order reaction has the rate constant, k = 4 . 6 × 10 ^ - 3...

For a first order reaction:
$t\dfrac{1}{2} = \dfrac{0 . 693}{k} = \dfrac{0 . 693}{4 . 6 \times {10}^{- 3}{s}^{- 1}} = 150 . 65s$

For a first order reaction,
{t}_{10 %} = \dfrac{1}{K}\ln \left(\dfrac{a}{a - x}\right) = \dfrac{1}{K}\ln \left(\dfrac{100}{90}\right)

{t}_{10 %} = \dfrac{2 . 303}{K} ( \log 10 - \log 9 )

{t}_{10 %} = \dfrac{2 . 093}{K} \times ( 0 . 04 )

Similarly

{t}_{90 %} = \dfrac{1}{K}\ln \left(\dfrac{100}{10}\right)

{t}_{90 %} = \dfrac{2 . 303}{K}

\dfrac{{t}_{90 %}}{{t}_{10 %}} = \dfrac{1}{0 . 04} = 25

We know,
${e}^{kt} = \dfrac{a}{a - x}$

$\dfrac{a - x}{a} = {e}^{- kt}$

$x = a\left(1 - {e}^{- kt}\right)$

$\alpha = \dfrac{x}{a} = \left(1 - {e}^{- kt}\right)$

Thus, only option (D) is correct.