A reaction proceeds by first order, 75% of this reaction was completed in 32 min. The time required for 50% completion is 

Given 75% reaction gets completed in 32 min

Thus, $k=\frac{2.303}{t}\log \frac{a}{(a-x)}=\frac{2.303}{32}\log \frac{100}{(100-75)}$

$=\frac{2.303}{32}\log 4=0.0433{\min }^{-1}$

Now we can use this value of k to get the value of time required for 50% completion of reaction

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}=\frac{2.303}{0.0433}\log \frac{100}{50}$

$=\frac{2.303}{0.0433}\log 2=16\min$