Answer
$$L=\frac{\sqrt{1+(\ln 5)^{2}}}{\ln 5}\left(5^{2 \pi}-1\right)$$
Work Step by Step
Given
$$
r=5^{\theta}, \quad 0 \leqslant \theta \leqslant 2 \pi
$$
The length is given by
\begin{align*}
L&=\int_{a}^{b} \sqrt{r^{2}+\left(\frac{d r}{ d \theta}\right)^{2}} d \theta\\
&=\int_{0}^{2 \pi} \sqrt{\left(5^{\theta}\right)^{2}+\left(5^{\theta} \ln 5\right)^{2}} d \theta\\
&=\int_{0}^{2 \pi} \sqrt{5^{2 \theta}\left[1+(\ln 5)^{2}\right]} d \theta\\
&=\sqrt{1+(\ln 5)^{2}} \int_{0}^{2 \pi} 5^{\theta} d \theta\\
&=\sqrt{1+(\ln 5)^{2}} \frac{5^{\theta}}{\ln 5}\bigg|_{0}^{2 \pi}\\
&=\sqrt{1+(\ln 5)^{2}}\left(\frac{5^{2 \pi}}{\ln 5}-\frac{1}{\ln 5}\right)\\
&=\frac{\sqrt{1+(\ln 5)^{2}}}{\ln 5}\left(5^{2 \pi}-1\right)
\end{align*}
