It is given that $x=2, y=3$

By the expansion of $(2x-3y{)}^8$ middle term is $5^{\text{th }}$ term.

$a={}^{8}C_4\left(2x{)}^4\right(3y{)}^4(-1{)}^4$

$\Rightarrow a=\frac{8!}{4!4!}\times 2^4\times 3^4\times x^2{y}^2$

$=70\times 2^4\times 3^4\times 2^4\times 3^4$

Similarly, in the expansion of $(3x+4y{)}^7, 4$th and $5$$\mathrm{th}$ terms are middle terms.

$b={}^{7}C_3\left(3x{)}^4\right(4y{)}^3$

$=35\times 3^7\times 2^{10}$

$c={}^{7}C_4\left(3x{)}^3\right(4y{)}^4$

$=35\times 3^7\times 2^{11}$

Now, $\frac{b+c}{2}=\frac{35\times 3^7\times 2^{10}+35\times 3^7\times 2^{11}}{70\times 2^4\times 3^4\times 2^4\times 3^4}$
$\Rightarrow \frac{b+c}{2}=\frac{2^{10}}{2^9}=2$