Solve Time Complexity problem using Time Hierarchy
Your intuition that $ \text{TIME}(n^3 \times 2^n) \subsetneq \text{TIME}(2^{n^2})$ is correct.

It is easy to see that $2^{n^2}$ is time-constructible. Given $n$, you can compute $n^2$ in $\log^2 n$ time, then write 100...00 with $n^2$ zeros, in time $O(n^2)$.

The Time-Hierarchy Theorem states that, for such time-constructible  $f(n)$,

$$ \text{TIME}(o(\frac{f(n)}{\log f(n)})) \subsetneq \text{TIME}(f(n))$$

It is clear that $ n^3 \times 2^n \in o(\frac{2^{n^2}}{\log 2^{n^2}}) = o(\frac{2^{n^2}}{n^2}) $.

Therefore, $ \text{TIME}(n^3 \times 2^n) \subseteq \text{TIME}(o(\frac{2^{n^2}}{\log 2^{n^2}})) \subsetneq \text{TIME}(2^{n^2})$.