The standard error for this test is
\begin{equation*}
\sigma_e = \frac{15}{\sqrt{49}} = \frac{15}{7}
\end{equation*}
and so using the t-distribution with degrees of freedom n-1 = 48 yields a t-statistic of
\begin{equation*}
t = \frac{\overline{x} - 200}{\sigma_e} = \frac{206-200}{\frac{15}{7}} = \frac{14}{5} = 2.80.
\end{equation*}
To compute the p-value,
\begin{equation*}
P(t \gt 2.80) + P(t \lt 2.80) \approx 0.0037 + 0.0037 = 0.0074.
\end{equation*}
Since this p-value is less than our significance level \(\alpha = 0.01\) then you can reject the null hypothesis and accept the alternate.