The correct answer is $\boxed{1 + x + \frac{3}{2}x^2 + x^3}$.
The Taylor series of a function $f(x)$ centered at $x=0$ is given by
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$
where $f^{(n)}(0)$ is the $n$th derivative of $f$ evaluated at $x=0$.
In this case, $f(x) = ex + x^2$,
so$$f'(x) = e^x + 2x$$
$$f”(x) = e^x + 2$$
$$f”'(x) = e^x$$
and so on.
Evaluating these derivatives at $x=0$ gives
$$f^{(0)}(0) = 1$$
$$f^{(1)}(0) = 1 + 2 = 3$$
$$f^{(2)}(0) = 1 + 2 = 3$$
$$f^{(3)}(0) = 1$$
Substituting these values into the Taylor series gives
$$f(x) = 1 + x + \frac{3}{2}x^2 + x^3$$
as desired.
The other options are incorrect because they do not include all powers of $x$ less than or equal to 3.