Math, as promised.

Suppose that the difference in utility is uniformly distributed,

U(b) - U(a) ~ Uniform(u,v)

Assume for simplicity that U(a)=0 and that E[U(b)] > 0, so that b is the better choice if there is no more information.

E[U(optimal|noinfo)] = E[U(b)] = (u+v)/2
E[U(optimal|info)] = integral_u^v dx if x<0 then 0 else x
= if 0 <= u <= v then (u+v)/2
if u <= 0 <= v then (0+v)/(v-u)*(0+v)/2 = v^2/2(v-u)

So, if u<0, you should pay at most (u^2 - 2v^2)/(2v - 2u) for information on whether U(b)>U(a).

If the difference is normally distributed with mean m and standard deviation s.

U(b) - U(a) = U(b) ~ Normal(m,s)

Then

E(U|no info) = E[U(b)] = m
E(U|info) = -- thank you, mathematica
Assuming[s > 0, Integrate[x PDF[NormalDistribution[m, s], x], {x, 0, Infinity}]]
= 1/2 (m + Exp[-m^2/(2 s^2)] Sqrt[2/pi] s + m Erf[m/(Sqrt[2] s)])
= s*normpdf(m/s) + m*normcdf(m/s)

A reasonable opproximation seems to be

E[U|info) ~= 0.4 s Exp[-2 (m/s)] + m

So, you should be willing to pay 0.4sExp[-2 (m/s)]. That means that you should pay exponentially less for each standard deviation that the mean is greater than 0. When the mean difference is 0, so when both are apriori equally likely, the information is worth s/sqrt(2pi) ~= 0.4 s. When the mean difference is one standard deviation in favor of b, the information is only worth 0.0833155 s.

To summarize: the more sure you are of which choice is best, the less the information that tells you that for certain is worth.