Answer: \(\alpha=1.89\) radians, to 3 significant figures.

For a polar curve \(r=f(\theta)\), the arc length from \(\theta=0\) to \(\theta=\alpha\) is

\(s=\int_0^\alpha \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta\).

Here \(r=e^{4\theta}\), so \(\frac{dr}{d\theta}=4e^{4\theta}\).

Hence

\(s=\int_0^\alpha \sqrt{e^{8\theta}+16e^{8\theta}}\,d\theta=\int_0^\alpha \sqrt{17}\,e^{4\theta}\,d\theta\).

Therefore

\(s=\left[\frac{\sqrt{17}}4e^{4\theta}\right]_0^\alpha=\frac{\sqrt{17}}4(e^{4\alpha}-1)\).

The length is 2015, so

\(\frac{\sqrt{17}}4(e^{4\alpha}-1)=2015\).

Thus

\(e^{4\alpha}=1+\frac{8060}{\sqrt{17}}=1955.837\ldots\).

Taking natural logarithms gives

\(\alpha=\frac14\ln(1955.837\ldots)=1.89\) radians, to 3 significant figures.