The confidence interval for the population mean at a [tex]100\times(1-\alpha)\%[/tex] confidence level is [tex]\mu\pm Z_{\alpha/2}\sigma_\mu[/tex]. Here [tex]\mu=50[/tex] is the sample mean, [tex]Z_{\alpha/2}[/tex] is the critical value such that [tex]\mathbb P(|Z|<Z_{\alpha/2})=1-\alpha[/tex], and [tex]\sigma_\mu=\dfrac\sigma{\sqrt n}[/tex], where in turn [tex]\sigma=6[/tex] is the assumed population standard deviation and [tex]n=100[/tex] 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 [tex]Z_{\alpha/2}[/tex] to lie somewhere between 2 and 3. To be more precise, you can consult a z-score table to find that [tex]Z_{\alpha/2}\approx2.3264[/tex].
Plug in everything you know and you find that the 98% confidence interval is
[tex]\left(50-2.3264\times\dfrac6{\sqrt{100}},50+2.3264\times\dfrac6{\sqrt{100}}\right)=(48.6042,51.3958)[/tex]
