1. Write the first-order decay expressions

For a first-order reaction, $[X]_t = [X]_0 e^{-k t}$

So for substances $A$ and $B$: $[A]_t = [A]_0 e^{-k_A t}, \qquad [B]_t = [B]_0 e^{-k_B t}$

Since the initial mixture is equimolar, $[A]_0 = [B]_0$

Hence, $\frac{[A]_t}{[B]_t} = \frac{[A]_0 e^{-k_A t}}{[B]_0 e^{-k_B t}} = e^{(k_B-k_A)t}$

We are given that at some time, $[A]_t = 16 [B]_t$ So, $\frac{[A]_t}{[B]_t} = 16$ Therefore, $e^{(k_B-k_A)t} = 16$

2. Calculate the rate constants from half-lives

For first-order kinetics, $k = \frac{\ln 2}{t_{1/2}}$

For $A$: $k_A = \frac{\ln 2}{54}$

For $B$: $k_B = \frac{\ln 2}{18}$

Thus, $k_B - k_A = \ln 2\left(\frac{1}{18} - \frac{1}{54}\right)$

Take LCM: $\frac{1}{18} - \frac{1}{54} = \frac{3-1}{54} = \frac{2}{54} = \frac{1}{27}$

So, $k_B-k_A = \frac{\ln 2}{27}$

3. Substitute into the ratio equation

$e^{(k_B-k_A)t} = 16$ $e^{\left(\frac{\ln 2}{27}\right)t} = 16$

Now, $16 = 2^4 = e^{4\ln 2}$

Therefore, $\frac{\ln 2}{27}t = 4\ln 2$

Cancel $\ln 2$: $\frac{t}{27} = 4$ $t = 108 \text{ min}$

4. Nearest integer

$t = 108 \text{ min}$

So the required integer answer is 108.