Given, 

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

Using the vector triple cross product we have,

$\overrightarrow{\mathrm{a}}\times \left(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right)=\left(\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{c}}\right)\overrightarrow{\mathrm{b}}-\left(\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{b}}\right)\overrightarrow{\mathrm{c}}$

So, $(\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{c}})\overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{b}})\overrightarrow{\mathrm{c}}=\frac{\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}}{2}$

On comparing both sides we get,

$\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{c}}=\frac{1}{2},\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{b}}=\frac{1}{2}$

$\because \overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{d}} = \overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{b}}$

$\therefore \overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{d}}=\frac{1}{2}$

Using vector cross-product formulae we have

$(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})·(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{d}})=\overrightarrow{\mathrm{a}}·(\overrightarrow{\mathrm{b}}\times (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{d}}\left)\right)$

$=\overrightarrow{\mathrm{a}}·\left(\right(\overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{d}})\overrightarrow{\mathrm{c}}-(\overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{c}}\left)\overrightarrow{\mathrm{d}}\right)$      $\left(\because \overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{c}}=0\right)$

$=(\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{c}})(\overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{d}})=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$