Given, $\overrightarrow{a}=2\hat{i}-\hat{j}+5\hat{k},\overrightarrow{b}=\alpha \hat{i}+\beta \hat{j}+2\hat{k}$

Also given $\left(\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \hat{i}\right)·\hat{k}=\frac{23}{2}$, then $\left|\overrightarrow{b}\times 2\hat{j}\right|$ is

Now using triple cross product we have, $\left(\left(\overrightarrow{a}·\hat{i}\right)\overrightarrow{b}-\left(\overrightarrow{b}·\hat{i}\right)\overrightarrow{a}\right)·\hat{k}=\frac{23}{2}$

$\Rightarrow \left(2.\overrightarrow{b}-\alpha ·\overrightarrow{a}\right).\hat{k}=\frac{23}{2}$

$\Rightarrow 2\times 2-\alpha \times 5=\frac{23}{2}$

$\Rightarrow 5\alpha =4-\frac{23}{2}$

$\Rightarrow \alpha =\frac{-3}{2}$

Now finding, $\overrightarrow{b}\times 2\hat{j}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \alpha & \beta & 2\\ 0& 2& 0\end{array}\right|=-4\hat{i}+2\alpha \hat{k}$

$\therefore \left|\overrightarrow{b}\times 2\hat{j}\right|=\sqrt{16+4{\alpha }^2}=\sqrt{16+4\times \frac{9}{4}}=5$