The confidence interval for the population mean at a \(100\times(1-\alpha)\%\) confidence level is \(\mu\pm Z_{\alpha/2}\sigma_\mu\). Here \(\mu=50\) is the sample mean, \(Z_{\alpha/2}\) is the critical value such that \(\mathbb P(|Z|, and \(\sigma_\mu=\dfrac\sigma{\sqrt n}\), where in turn \(\sigma=6\) is the assumed population standard deviation and \(n=100\) is the sample size.
First find the critical value. By the empirical rule, you might already know that about 95% of a normal distribution lies within two standard deviations of the mean, and that about 99.7% lies within three standard deviations, so you can expect \(Z_{\alpha/2}\) to lie somewhere between 2 and 3. To be more precise, you can consult a z-score table to find that \(Z_{\alpha/2}\approx2.3264\).
Plug in everything you know and you find that the 98% confidence interval is
\(\left(50-2.3264\times\dfrac6{\sqrt{100}},50+2.3264\times\dfrac6{\sqrt{100}}\right)=(48.6042,51.3958)\)
