Answer
2036.8
Work Step by Step
Given $\mu=2000,\sigma=100,n=20$
the sampling distribution will be normal with a mean
of $2000$ and standard error of $\frac{100}{\sqrt {20}}=22.36$
For $P(z)=0.95$, use table E or other sources, we can get $z=1.645$
thus $\frac{X-2000}{22.36}=1.645$ and we find $X=2036.8$
The sample mean that will cut off the upper 95% of all
samples of size 20 taken from the population is 2036.8
