Work this out from first principles using the calcium that actually got used up.
The 100 ml of gypsum solution supplied calcium at $36$ meq/l, so the total calcium put into the test was $36 imes 0.1 = 3.6$ meq.
After reacting with the 5 g soil sample, the filtrate still carried $32$ meq/l of Ca+Mg, meaning $32 imes 0.1 = 3.2$ meq of that calcium and magnesium remained unused.
So the calcium consumed by the soil, which is the actual gypsum requirement of the 5 g sample, is $3.6 - 3.2 = 0.4$ meq.
Scaling this 5 g result up to a 100 g soil basis, the requirement becomes $0.4 imes rac{100}{5} = 8$ meq per 100 g of soil.
\[ oxed{8\ ext{meq}/100\ ext{g soil}} \]