Step 1: Recognize the density as a Gamma distribution.
The density $f(x)=4x^2e^{-2x}$ for $x>0$ is a Gamma density with shape $\alpha=3$ and rate $\beta=2$, since the normalizing constant $\dfrac{\beta^\alpha}{\Gamma(\alpha)}=\dfrac{2^3}{2!}=4$ matches.
Step 2: Recall the formula for the mean log of a Gamma variable.
For $X\sim Gamma(\alpha,\beta)$, there is a standard result using the digamma function $\psi$: \[ E[\ln X]=\psi(\alpha)-\ln\beta \] For an integer shape $\alpha=3$, the digamma function has the closed form \[ \psi(3)=-\gamma+1+\frac{1}{2}=\frac{3}{2}-\gamma \] where $\gamma\approx0.5772$ is the Euler-Mascheroni constant.
Step 3: Plug in the numbers.
\[ \psi(3)=1.5-0.5772=0.9228 \] \[ \ln\beta=\ln2=0.6931 \] \[ E[\ln X]=0.9228-0.6931=0.2297 \] So $E(Y)\approx0.2297$, a finite number. This already tells us option (A) is wrong.
Step 4: Compute the benchmark value $\ln(3/2)$.
\[ \ln\frac{3}{2}=\ln1.5\approx0.4055 \]
Step 5: Compare.
Since $0.2297<0.4055$, we get $E(Y)<\ln(3/2)$ strictly. This rules out (B), which needs exact equality, and (C), which needs the opposite direction.
Step 6: Cross check with the concavity idea.
This makes sense because $\ln$ bends downward, so averaging $x$ first and then taking the log always gives a bigger number than taking the log first and then averaging, as long as $X$ is not a fixed constant. Here $E(X)=\alpha/\beta=3/2$, so $\ln E(X)=\ln(3/2)$ sits above $E(\ln X)$, matching our direct calculation. This concavity gap is not a small rounding effect either: the difference $\ln(3/2)-E[\ln X]\approx0.4055-0.2297\approx0.1758$ is a genuine, strictly positive gap. Since $X$ has a proper spread coming from its Gamma(3,2) density, that gap can never close.
Step 7: Conclude.
\[ \boxed{E(Y)<\ln\dfrac{3}{2}} \]