Let a tangent on parabola $y^2=16x$ be,

$y=mx+\frac{4}{ m}$

It is also common to $x^2+y^2=8$

$\text{If} {\left(\frac{4}{m}\right)}^2=8\left(1+m^2\right)$

$\Rightarrow \frac{16}{m^2}=8(1+m^2)$

$\Rightarrow m^4+m^2-2=0\Rightarrow (m^2+2)(m^2-1)=0$

$\Rightarrow m=\pm 1$

Taking one of the tangents $y=x+4$

$\Rightarrow x-y+4=0$

Tangent to $x^2+y^2=8$ is $x{x}_1+y{y}_1-8=0$

$\therefore \frac{x_1}{1}=\frac{y_1}{1}=\frac{-8}{4}$

$\Rightarrow Q(-2,2)$

Tangent to $y^2=16x$ is $y{y}_1=8(x+x_1)$

$\Rightarrow 8x-y{y}_1+8x_1=0$

$\therefore \frac{8}{1}=\frac{-y_1}{-1}=\frac{8x_1}{4}$

$\Rightarrow y_1=8, x_1=4$

$\therefore R(4,8)$

Using the distance formula between two points,

$QR=\sqrt{(4+2{)}^2+(8-2{)}^2}=6\sqrt{2}$

$\therefore (QR{)}^2={\left(6\sqrt{2}\right)}^2=72$