If the fraction of glucose present as the alphaanomer is a the fraction present as the $\beta$ anomer is b, and the rotation of the mixture is +52.6 degree, we have
a(+112.2degree) + b(18.7) =52.6 degree
There is a very little of the open chain form present, so the fraction present as the $\alpha$ anomer (a) plus the fraction present as the $\beta$ anomer (b) should account for all the glucose a+b = 1 or b = 1-a.
Thus 1(+112.2degree) + (1-a)(18.7) =52.6 degree
solving this equation for (a), we have a = .36 or 36 percentage. Thus (b) must be (1-0.36) = .64 or 64 percentage
Hence (a) is the correct answer.