Using spherical coordinates, with equal to the radius of the sphere, 1, we can write the "position vector" of any point on the surface as
Differentiating with respect to and gives two vectors in the tangent plane.
The cross product of those two tangent vectors, the "fundamental vector product" for this surface, gives the "vector differential of surface area",
(The order in which you take the cross product determines the sign. I chose the way that gives an outward pointing normal [all components positive] because the problem said "the outward flux".)
You want to integrate over the surface of the sphere: from 0 to , from 0 to . Of course, you will need to change to these coordinates.