If $z=\cos 6°+i\sin 6°,$ then $\sum _{n=1}^{20}Im\left(z^{2n-1}\right)=$

$\sum _{n=1}^{20}{z}^{2n-1}=z+z^3+z^5+\dots +z^{39}$
Now, $z=\cos 6°+i\sin 6°=e^{i6°}$

Substitute the value in the series,

$\sum _{n=1}^{20}{z}^{2n-1}=e^{i6°}+e^{i18°}+e^{i30°}+\dots +e^{i\left(39\times 6°\right)}$

$=e^{i6°}\left(1+e^{i{12}^0}+e^{i{24}^0}+\dots \dots +e^{i{228}^0}\right)$

$=e^{i6°}\left(1+e^{i{12}^0}+e^{i{24}^0}+\dots \dots +e^{i{228}^0}\right)$

$=e^{i6°}\left[\frac{e^{i{240}^0}-1}{e^{i{12}^0}-1}\right]$
Simplify the above equation

$\sum _{n=1}^{20}Im\left(z^{2n-1}\right)=e^{i6°}\left[\frac{e^{i{240}^0}-1}{e^{i{12}^0}-1}\right]$

$=\frac{e^{i6°}\left[\frac{-1}{2}-\frac{\sqrt{3}}{2}i-1\right]}{e^{i6°}\left(2i\sin 6°\right)}$

.$=\frac{\left[-\frac{3}{2}-\frac{\sqrt{3}}{2}i\right]}{2i\sin 6°}$

$=-\frac{\sqrt{3}}{4\sin 6°}+\frac{3}{4\sin 6°}i$

Thus, $\sum _{n=1}^{20}Im\left(z^{2n-1}\right)=\frac{3}{4\sin 6°}$