[126.4, 132.2] is a 95% confidence interval for the mean μ of a normally distributed random variable with known variance. Find a 98% confidence interval for μ, based on the same sample.
I got gamma = 0.98 gamma/2 = 0.49 so Z* = 2.33 but then I am lost on what to do next.
The half-width of the $95\%$ CI you're given is $2.9.$ The half-width of a $95\%$ confidence interval for known variance $\sigma^2$ is $$ \frac{\sigma}{\sqrt n}(1.96)$$ so this means $\frac{\sigma}{\sqrt n} = \frac{2.9}{1.96}=1.48.$
As your calculation shows, the half-width of a $98\%$ CI is $$ \frac{\sigma}{\sqrt{n}}(2.33)$$ so the half-width of your $98\%$ CI must be $$1.48\times 2.33 =3.45.$$