The magnetic flux through an area A is given by

where
B is the magnitude of the magnetic field
A is the area

is the angle between the direction of B and the perpendicular to the surface A.
In our problem, the area lies in the x-y plane, while B is in the z direction, this means that B and the perpendicular to A are parallel, so

and

, so we can rewrite the formula as

We can calculate the area starting from the radius:

And then using the intensity of the magnetic field given by the problem,

, we find the magnetic flux:
