Answer
$\color{blue}{\sigma_C \approx 8.72^{\circ}\text{C}}$
Work Step by Step
For any given temperature $Y$ in the Fahrenheit scale, let
$\quad C = \frac{5}{9}(Y-32) = \frac{5}{9}Y- \frac{160}{9}$
be the equivalent temperature in the Celsius scale. We are given that
$\quad \sigma_Y = 15.7^\circ\text{F}$.
Since $\text{Var}(Y) = \sigma_Y^2 = (15.7^\circ\text{F})^2$ is finite, then $\mu$ exists (as $\text{Var}(Y) = E(Y^2) - \mu_Y^2$ by Theorem 3.6.1, p. 155) so that we can use Theorem 3.6.2 with $a = \frac{5}{9}$ and $b = -\frac{160}{9}$ to get
$\begin{align*}
\text{Var}\left(C\right) &= \text{Var}\left(\left(\frac{5}{9}Y + \left(-\frac{160}{9}\right) \right)\cdot \left(\frac{^\circ\text{C}}{^\circ\text{F}}\right)\right) \\
&= \left(\frac{5}{9} \cdot \left(\frac{^\circ\text{C}}{^\circ\text{F}}\right)\right)^2\cdot\text{Var}(Y) \qquad [\ \text{Theorem 3.6.2}\ ]\\
\text{Var}\left(C\right) &= \left(\frac{5}{9} \cdot \left(\frac{^\circ\text{C}}{^\circ\text{F}}\right)\right)^2\cdot (15.7^\circ\text{F})^2 \\
\sqrt{\text{Var}\left(C\right)} &= \frac{5}{9} \cdot (15.7^\circ\text{C}) \\
\color{blue}{\sigma_C}\ &\color{blue}{\approx 8.72^\circ\text{C}}
\end{align*}$
