I/I_{0} = 0.85

I_{0}/I = 1/0.85

A_{1}(for 1 cm) = log(I_{0}/I) = log(1/0.85) = 0.0706

Since A = e cl amd e
and c are constant then A_{2}(for 3 cm) = 3.0X0.0706

= 0.2117 = log(I_{0}/I_{2})

I_{0}/I_{2} = antilog 0.2117 = 1.628

%T = (I_{2}/I_{0})100 = 100/1.628 = 61.4%

Answer is B