Calculating phosphorescence quantum yield using the given datas'
The triplet can only be formed by intersystem crossing from the singlet. The fluorescence yield is $\phi_f=k_f^0/(k_f^0+k_i)$ where $k_i$ is the intersystem crossing rate constant to form the triplet and $k_f^0$ the radiative rate constant for fluorescence. The fluorescence lifetime is $\tau=1/(k_f^0+k_i)=10\cdot 10^{-9}$. The triplet yield is $1-\phi_f$. The triplet lifetime is not important and as the triplet only phosphoresces (you mention no other processes, so no, non-radiative, pathway from triplet) all the  molecules formed must phosphoresce (and it does not matter how long this takes as its the only pathway out of the triplet.)  However, you still have one piece of evidence missing, namely $k_f^0$.