How would you go about finding the mean cross-sectional area of a sphere that has a radius equal to $1$?
I have calculated the mean length of a chord in a circle (with radius $1$) to be $\frac{4}{\pi}$, so would I be able to use this answer to help me find the solution to this new problem?
Would I have to take into account that areas can be at varying angles, or could I assume it is the same as the mean of a set of all the areas that are parallel to each other?
Since you calculated the average chord in the 2D case by uniformly varying the subtended angle, I think it makes most sense to let the diameter of the cross section be defined by the chords on a great circle of the sphere, with the subtended angle of the chord varying uniformly from $0$ to $2\pi$
For a subtended angle $\theta$, the chord length / diameter is $d(\theta)=\sqrt{(1-\cos\theta)^2+(\sin\theta)^2}=\sqrt{2-2\cos\theta}$
The area of a circle of diameter $d$ is $A(d)=\dfrac{\pi d^2}{4}$
Thus the area as a function of $\theta$ is $A(\theta)=\frac{\pi}{4}(2-2\cos\theta)$
Finding the average value $\mu$ of this over all angles \begin{align} \mu&=\frac{1}{2\pi}\int_0^{2\pi}A(\theta)d\theta\\ &=\frac{1}{2\pi}\int_0^{2\pi}\frac{\pi}{4}(2-2\cos\theta)d\theta\\ &=\frac{1}{2\pi}\frac{\pi}{4}\left[2\theta-2\sin\theta\right]_0^{2\pi}\\ &=\frac{\pi}{2} \end{align} So with this setup, the mean cross sectional area of the unit sphere is $\dfrac{\pi}{2}$
In response to your final question about accounting for areas at different angles, notice that this result is equal regardless of which great circle we vary the chord / diameter around. Therefore, by symmetry, you do not need to consider other orientations that the cross section could vary about.