1. Since the half-lives are independent of initial concentration, both A and B undergo first-order decay.

2. Let the initial concentrations be equal: $[A]_0 = [B]_0 = C_0$

3. For a first-order reaction: $[X]_t = [X]_0 e^{-kt}$ and $k = \frac{\ln 2}{t_{1/2}}$

4. Calculate rate constants:

   For A: $k_A = \frac{0.693}{100} = 0.00693\ \text{s}^{-1}$

   For B: $k_B = \frac{0.693}{50} = 0.01386\ \text{s}^{-1}$

5. Concentrations after time $t$: $[A] = C_0 e^{-k_A t}$ $[B] = C_0 e^{-k_B t}$

6. Given that concentration of A becomes four times that of B: $[A] = 4[B]$

   Substituting: $C_0 e^{-k_A t} = 4C_0 e^{-k_B t}$

   Cancel $C_0$: $e^{(k_B-k_A)t} = 4$

7. Now, $k_B - k_A = 0.01386 - 0.00693 = 0.00693$

   So, $e^{0.00693t} = 4$

   Taking natural log: $0.00693t = \ln 4 = 2\ln 2 = 2(0.693)=1.386$

8. Hence, $t = \frac{1.386}{0.00693} = 200\ \text{s}$

Therefore, the required time is: $\boxed{200\ \text{s}}$