Given: The reaction $A \to B$ is zero order.

For a zero-order reaction, the integrated rate law is: $[A]_t = [A]_0 - kt$

The half-life $t_{1/2}$ is the time when: $[A]_t = \frac{[A]_0}{2}$ So, $\frac{[A]_0}{2} = [A]_0 - kt_{1/2}$ $kt_{1/2} = \frac{[A]_0}{2}$ $t_{1/2} = \frac{[A]_0}{2k}$

Given $t_{1/2} = 50$ min: $\frac{[A]_0}{2k} = 50$ $\frac{[A]_0}{k} = 100$

We need the time when concentration becomes one-fourth of initial value: $[A]_t = \frac{[A]_0}{4}$

Using zero-order law: $\frac{[A]_0}{4} = [A]_0 - kt$ $kt = [A]_0 - \frac{[A]_0}{4} = \frac{3[A]_0}{4}$ $t = \frac{3[A]_0}{4k}$

Using $\frac{[A]_0}{k} = 100$: $t = \frac{3}{4} \times 100 = 75 \text{ min}$

Final Answer: $\boxed{75}$