Answer: $k' =1.507 \times 10^{-5}M^{-1}s^{-1}$

Given $[A] = 55\;M$ at $39s$, $[A]_o = 99\;M$, and $[B]_o = 1000\;M$.

Since $[B]_0 \gt \gt [A]_0$ we can treat this as a pseudo-first order reaction.

We can use the rate equation $[A] = [A]_0 e^{-k't[B]_0}$

$\Rightarrow 55 = 99 e^{k' \times 39 \times 1000} \rightarrow k' =1.507 \times 10^{-5}M^{-1}s^{-1}$