Given,

$\overrightarrow{a}$ and $\overrightarrow{b}$ be two vectors such that $\left|\overrightarrow{a}\right|=1, \left|\overrightarrow{b}\right|=4$ and $\overrightarrow{a}\cdot \overrightarrow{b}=2$.

Also, given

$\overrightarrow{c}=\left(2\overrightarrow{a}\times \overrightarrow{b}\right)-3\overrightarrow{b}$

Taking dot product with $\overrightarrow{b}$ both side we get,

$\Rightarrow \overrightarrow{b}\cdot \overrightarrow{c}=\left(2\overrightarrow{a}\times \overrightarrow{b}\right)\cdot \overrightarrow{b}-3{\left|b\right|}^2$

$\Rightarrow \left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos \alpha =0-3{\left|\overrightarrow{b}\right|}^2$

$\Rightarrow \left|\overrightarrow{c}\right|\cos \alpha =-12$ {as $\left|\overrightarrow{b}\right|=4$}

Now, solving $\overrightarrow{a}\cdot \overrightarrow{b}=2$

$\Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta =2$

$\Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}$

Now, solving

${\left|\overrightarrow{c}\right|}^2={\left|\left(2\overrightarrow{a}\times \overrightarrow{b}\right)-3\overrightarrow{b}\right|}^2$

$\Rightarrow {\left|\overrightarrow{c}\right|}^2=64\times \frac{3}{4}+144=192$

Now, squaring both side of $\left|\overrightarrow{c}\right|\cos \alpha =-12$ we get,

${\left|\overrightarrow{c}\right|}^2{\cos }^2\alpha =144$

$\Rightarrow 192{\cos }^2\alpha =144$

$\Rightarrow 192\left(1-{\sin }^2\alpha \right)=144$

$\Rightarrow 192{\sin }^2\alpha =48$