In a first-order reaction, the concentration of the reactant decreases from 0.8 Μ to 0.4 Μ in 15 minutes. The time is taken for the concentration to change from 0.1 Μ to 0.025 Μ is:

30 minutes

The correct answer is option 1. 30 minutes. 

Order is one here as concentration changes from 0.8 Μ to 0.4 Μ (50%) in 15 minutes, this half-life is 15 minutes = \(T_{50}\)
A change from 0.1 Μ to 0.025 Μ is 75% and for a first-order reaction

\(T_{50} = 15 min\)

\(k = \frac{0.693}{T_{50}}\)

   \(k = 0.0462\)

\(\alpha = 0.1M\)

\((\alpha − x) = 0.025 M\)

For first-order reaction

\(k = \frac{2.303}{t}log\frac{\alpha}{\alpha − x}\)

\(⇒ 0.0462 = \frac{2.303}{t}log\frac{0.1}{0.025}\)

\(⇒ 0.0462 = \frac{2.303}{t}\times 0.602\)

\(⇒ t = \frac{2.303 × 0.602}{0.0462}\)

\(⇒ t \approx 30 min\)