Answer: 151 s
Step-1: Calculate $k$, given $t_{1/2}$:
The half life $t_{1/2} = \large\frac{ ln\;2}{k}$$ \rightarrow k = \large\frac{0.693}{t_{1/2}}$$ = \large\frac{0.693}{35}$$ = 0.0198 s^{-1}$
Step-2: Calculate the time required for 95% of phosphine to decompose
We know that $\log \large\frac{[A_t]}{[A_0]}$$ = - \large\frac{kt}{2.303}$
$\Rightarrow t = - \log \large\frac{0.5}{1}$$\times \large\frac{2.303}{0.0198}$$ = -1.3\times -\large\frac{ 2.303}{0.0198}$ $=151\;s$