let no. of femitted quanta = n1
& of absorbed = n2
frequency of emitted quanta  f1 =  c/ 5080* 10^-10
energy e1  =  h*f1 = h*  c/ 5080* 10^-10
frequency of absorbed quanta f2 = c/ 4530* 10^-10
energy e2 =   h*  c/ 4530* 10^-10
   total energy of emitted quanta =  0.47 0f  absorbed quanta
hence,
 n1* h*  c/ 5080* 10^-10 =   0.47*n2* h*  c/ 4530* 10^-10
n1/5080 = 0.47 * n2 /4530
n1/n2 =  0.527
						

							
Quantum efficiency(Q.E) is defined as ratio of quanta emitted to the number absorbed
it is expressed in ratio of energies
quanta absorbed on dye : E = hc/lamda = 1242 eV-nm /453 nm = 2.742 eV
quanta exhibited/undergo fluorescence on dye : E = hc/lamda = 1242 eV-nm /508 nm = 2.445 eV
 
Q.E = 0.47*2.742 / 2.445
       = 0.527
or 52.7 % efficient
I'm not sure if the method was correct, plz verify with someone else
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E(absorbed)=1242/453=2.742 
E(energy of photon)=1242/508=2.445
47% of absorbed energy is re-emitted out:
47/100*2.742
no. of photons emitted=energy beingemitted/energy of 1 photon
=47/100*2.742/2.444=0.527