Solution:

Let the bases be $b$ and $b + 100.$ The midsegment has length $b + 50$ and splits the trapezoid into two trapezoids of equal height, whose areas are proportional to the sums of their parallel sides, $b + (b + 50)$ and $(b + 50) + (b + 100).$ Setting $\frac{2b + 50}{2b + 150} = \frac{2}{3}$ gives $b = 75,$ so the bases are $75$ and $175.$

Extend the legs to meet at an apex, creating similar triangles: a segment parallel to the bases with length $\ell$ cuts off a triangle of area $c\ell^2$ for a fixed constant $c.$ The segment of length $x$ bisects the trapezoid's area exactly when $cx^2 - c \cdot 75^2 = c \cdot 175^2 - cx^2,$ so $$x^2 = \frac{75^2 + 175^2}{2} = 18125.$$

Then $x^2/100 = 181.25,$ and the greatest integer not exceeding it is $181.$