Given,

$A, B \& C$ are independent so,

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

$P\left(B\cap B\right)=P\left(B\right)P\left(C\right)$

$P\left(C\cap A\right)=P\left(C\right)P\left(A\right)$

And given $P\left(\overline{B}\cup \overline{C}\right)=\frac{1}{2}$

Now,

$P\left(\frac{\overline{B}\cap \overline{C}}{A}\right)=\frac{P\left(\left(\overline{B}\cap \overline{C}\right)\cap A\right)}{P\left(A\right)}$

By diagram we can see,

$P\left(\left(\overline{B}\cap \overline{C}\right)\cap A\right)=P\left(A\right)-P\left(A\cap B\right)-P\left(A\cap C\right)+P\left(A\cap B\cap C\right)$

Also, $P\left(\overline{B}\cup \overline{C}\right)=\frac{1}{2}$

$\Rightarrow 1-P\left(B\cap C\right)=\frac{1}{2}$

$\Rightarrow P\left(B\right)\left(P\left(C\right)\right)=\frac{1}{2}$

Rewriting,

$P\left(\left(\overline{B}\cap \overline{C}\right)\cap A\right)=P\left(A\right)-P\left(A\right)P\left(B\right)-P\left(A\right)P\left(C\right)+P\left(A\right)P\left(B\right)P\left(C\right)$

$\Rightarrow P\left(\left(\overline{B}\cap \overline{C}\right)\cap A\right)=P\left(A\right)\left[1-P\left(B\right)-P\left(C\right)+\frac{1}{2}\right]$

$\Rightarrow \frac{P\left(\left(\overline{B}\cap \overline{C}\right)\cap A\right)}{P\left(A\right)}=\left[1-P\left(B\right)-P\left(C\right)+\frac{1}{2}\right]$

$\Rightarrow \frac{P\left(\left(\overline{B}\cap \overline{C}\right)\cap A\right)}{P\left(A\right)}=\left(\frac{3}{2}-b-C\right)$

So, $P\left(\frac{\overline{B}\cap \overline{C}}{A}\right)=\frac{3}{2}-b-C$