For a first-order reaction, the rate constant $k$ is related to the half-life $t_{1/2}$ by: $$ \begin{aligned} k &= \frac{\ln 2}{t_{1/2}} \\ &= \frac{0.693}{50 \text{ s}} \\ &= 0.01386 \text{ s}^{-1} \end{aligned} $$ The time $t$ required for a first-order reaction to reach a certain completion is given by: $$ \begin{aligned} t &= \frac{1}{k} \ln\left(\frac{A_0}{A_t}\right) \end{aligned} $$ For $75\%$ completion, the remaining amount $A_t = A_0 - 0.75 A_0 = 0.25 A_0$. Thus, $\frac{A_0}{A_t} = \frac{1}{0.25} = 4$. $$ \begin{aligned} t &= \frac{1}{0.01386} \ln(4) \\ &= \frac{1.386}{0.01386} \\ &= 100 \text{ s} \end{aligned} $$ It will take $100$ seconds to complete $75\%$ of the reaction.