Surface area is given by
$ \iint_R \left| \vec r_u \times \vec r_v \right| \ dA $
where $\vec r(u,v)$ is the parametrization of the surface. We can rewrite this as (derivation shown here: http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx):
$ \iint_D \sqrt{ \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 + 1} \ dA $
for a function $z = f(x,y)$ where $D$ is the projection of the surface onto the xy-plane.
Since we are only concerned with the portion of the unit sphere above $z = 0$, we can write it as
$ z = \sqrt{1-x^2-y^2} $
Computing the partial derivatives with respect to $x$ and $y$,
$ \frac{\partial z}{\partial x} = \frac{-x}{\sqrt{1-x^2-y^2}} \rightarrow \left(\frac{\partial z}{\partial x}\right)^2 = \frac{x^2}{1-x^2-y^2} $
$ \frac{\partial z}{\partial y} = \frac{-y}{\sqrt{1-x^2-y^2}} \rightarrow \left(\frac{\partial z}{\partial y}\right)^2 = \frac{y^2}{1-x^2-y^2} $
Substituting these into our expression for surface area,
$ \iint_D \sqrt{ \frac{x^2}{1-x^2-y^2} + \frac{y^2}{1-x^2-y^2} + 1} \ dA $
which simplifies to (omitting a bit of algebra)
$ \iint_D \frac{1}{\sqrt{1-x^2-y^2}} \ dA $
Observe that $D$ (the projection of our surface into the xy-plane) is given by
$ z = \sqrt{1-x^2-y^2} $
$ \frac{1}{2} = \sqrt{1-x^2-y^2} $
$ \frac{1}{4} = 1-x^2-y^2 $
$ x^2+y^2 = \frac{3}{4} $
which is a circle of radius $\frac{\sqrt{3}}{2}$. The integral over $D$ is easiest done in polar coordinates. I'll assume you know how to do that and omit the computation.
$ \int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{3}}{2}} \frac{1}{\sqrt{1-r^2}} \ r \ dr \ d\theta $
$ = \pi $