Answer
$23^{\circ} \mathrm{C}$
Work Step by Step
Given \begin{equation}
c=340.3 \sqrt{\frac{T+273.15}{288.15}}.
\end{equation} Set $c = 345$ to find the temperature at sea level.
\begin{equation}
\begin{aligned}
345&=340.3 \sqrt{\frac{T+273.15}{288.15}}\\
\frac{345}{340.3}&=\sqrt{\frac{T+273.15}{288.15}}\\
(1.013811)^2& =\left( \sqrt{\frac{T+273.15}{288.15}}\right)^2\\
1.0278134& = \frac{T+273.15}{288.15}\\
288.15\cdot 1.0278134-273.15&\approx T\\
23& \approx T.
\end{aligned}
\end{equation} The temperature at sea level is about $23^{\circ} \mathrm{C}$.
