At 25 m, L = 80dB = 10 log(I/I_{0})

I/I_{0} = 10^{8}

I = 10^{8 }I_{0}= I_{1}

(I_{1}/I_{2})_{=}(r_{1}/r_{2})^{2}

I_{2} = I_{1}(r_{1}/r_{2})^{2} = (I_{0}10^{8})(25/50)^{2} = I_{0}10^{8}/4

Therefore at 50 m, L = 10 log[(1/4)I_{0}10^{8}/I_{0}] = 10 log 10^{8} - 10 log 4 = 80 - 6 = 74 dB

Answer is E