Write $w=\frac{4z^2+3z+2}{4z^2-3z+2}=1+\frac{6z}{4z^2-3z+2}=1+\frac{6}{2\left(2z+\frac{1}{z}\right)-3}$ after dividing numerator and denominator of the second fraction by $z$.

For $w\in \mathbb{R}$, the quantity $2z+\frac{1}{z}$ must be real. Equating it to its conjugate gives $2z+\frac{1}{z}=2\bar{z}+\frac{1}{\bar{z}}$.

Rearranging, $2(z-\bar{z})-\frac{z-\bar{z}}{|z{|}^2}=0$, i.e., $(z-\bar{z})\left(2-\frac{1}{|z{|}^2}\right)=0$.

Since the imaginary part of $z$ is non-zero, $z\ne \bar{z}$. Therefore $2-\frac{1}{|z{|}^2}=0$, which gives $|z{|}^2=\frac{1}{2}=0.50$.