Using $\sin \theta = \sin (\pi - \theta),$ $\displaystyle \prod_{k=1}^{10} \sin^2 \left( \dfrac{k \pi}{21} \right) = \displaystyle \prod_{k=1}^{20} \sin \left( \dfrac{k \pi}{21} \right).$ Let $\omega = e^{i \pi / 21}$ be the complex root of $\omega^{21} = 1.$ Using $\sin \theta = \dfrac{e^{i \theta} - e^{- i \theta}}{2},$ the desired product is equal to $\begin{aligned} \displaystyle \prod_{k=1}^{20} \dfrac{\omega^k - \omega^{-k}}{2i} & = \left( \displaystyle \prod_{k=1}^{20} 1 - \omega^{-k} \right) \omega^{210} \cdot \dfrac{1}{i^{20}} \cdot \dfrac{1}{2^{20}} \\ & = \dfrac{1}{2^{20}} \displaystyle \prod_{k=1}^{20} \left( 1 - \omega^{-k} \right) \\ & = \dfrac{1}{2^{20}} \displaystyle \sum_{k=0}^{n} 1^k \\ & = \dfrac{21}{2^{20}}. \end{aligned}$ The last step follows from the fact that the polynomial $f(x) = \displaystyle \prod_{k=1}^{20} (x - \omega^{-k} )$ has roots $e^{i \pi /21}, e^{2i \pi / 21}, \cdots , e^{20 i \pi / 21},$ so it must be identically equal to $x^{20} + x^{19} + \cdots + 1.$

some hints for those who don't know complex no.s

use sin(60-x)sin(60+x)sin(x)=0.25*sin(3x)

later form cubic equation for sine by using 4x=180-3x

check the product of roots

put the values to get the answer