Sample size for equal type I and II error of 5 percent

A sales manager wants to detect a one-deal shift in the weekly mean number of deals per salesperson. He tests $H_0:\mu=15$ against $H_a:\mu=16$ with known variance $\sigma^2=6$, assuming normally distributed counts and simple random sampling. How many observations are needed so that both the significance level and the type II error probability equal $0.05$?

$65$. For a one-sided z-test, $n=\dfrac{(z_\alpha+z_\beta)^2\sigma^2}{(\mu_a-\mu_0)^2}=\dfrac{(2\times1.645)^2\times 6}{1}\approx64.9$, which rounds up to $65$.