A reaction of first-order completed 90% in 90 minutes, hence, it is completed 50% in approximately :

Given, reaction of first-order completed 90% in 90 minutes.

The time required for 50% competition,

$$\begin{gathered} \frac{\ln 2}{t_{1/2}}=\frac{1}{90}\ln \left(\frac{100}{10}\right) \\ \Rightarrow t_{1/2}=27\min \end{gathered}$$