For a first order reaction A → B , the rate constant, k = 5 . 5 ×...

For first order reaction, $t = \dfrac{2 . 303}{k}\log \left[\dfrac{{A}_{0}}{{A}_{t}}\right]$

Now, {t}_{67 %} = \dfrac{2 . 303}{k}\log \left[\dfrac{A}{0 . 33A}\right]

& ${t}_{50} = \dfrac{2 . 303}{k}\log \left[\dfrac{A}{0 . 5A}\right]$

{t}_{67 %} = \dfrac{1}{k}\ln \left(\dfrac{1}{1 - 0 . 67}\right) = \dfrac{{t}_{1/2}}{\ln 2} \times \ln \left(\dfrac{1}{1 - \dfrac{2}{3}}\right)

⇒ {t}_{67 %} = 1 . 585 \times {t}_{1/2}

$X \times {10}^{- 1} = 1 . 585$

$⇒ X = 15 . 85$