Answer
The net magnetic flux is $0.040~weber$ (into the page)
Work Step by Step
We can find the magnetic flux through the left half of the loop:
$\phi = B~A$
$\phi = (2.0~T)(0.20~m)(0.20~m)$
$\phi = 0.080~weber$ (into the page)
We can find the magnetic flux through the right half of the loop:
$\phi = B~A$
$\phi = (1.0~T)(0.20~m)(0.20~m)$
$\phi = 0.040~weber$ (out of the page)
The net magnetic flux is $0.040~weber$ (into the page)
