Photochemistry and steady state approximation
The processes are absorb light into A, and then A* has only two pathways, i.e. fluoresces and reacts with B to form B*.

The B* has only two pathways, it  fluoresces and converts to C. All B* molecules must therefore follow one of these paths, so the sum of yields of B* (paths fluorescence + forming C) must be the same as the yield to form B* in the first place, thus you don't need to do a calculation.

In general if you have to find a yield of any process x this is, $\displaystyle \phi_x = \frac{\text{rate of process x}}{\text{rate of absorption}} \equiv \frac{\text{rate const x} }{\text{sum of all other rate constants from x}}$. 

In your example $\displaystyle \phi_{ET}= \frac{k_{ET}[B]}{k_f^A+k_{ET}[B]}$ where superscript $A$ refers to fluorescence from $A^*$.

To work all this out write down $d[A^*]/dt = +k_{abs}[A]- k_f^A[A^*]-k_{ET}[A^*][B]$ and use steady state.