Let $A = [a, \infty)$ denotes the domain, then $f: [a, \infty) \to B$, which is defined by $f(x) = 2x^3 - 3x^2 + 6$ will have an inverse for the smallest real value of 'a' if

Given $f(x) = 2x^3 - 3x^2 + 6$

Differentiating with respect to $x$, we get:

$f'(x) = 6x^2 - 6x = 6x(x - 1)$

For $f(x)$ to have an inverse, it must be bijective, which means it must be strictly monotonic (either strictly increasing or strictly decreasing) on its domain $[a, \infty)$.

The critical points are $x = 0$ and $x = 1$.

$f'(x) > 0$ for $x \in (-\infty, 0) \cup (1, \infty)$

$f'(x) < 0$ for $x \in (0, 1)$

For $f(x)$ to be strictly increasing on $[a, \infty)$, we must have $a \ge 1$. The smallest real value of $a$ for which $f(x)$ is one-one is $a = 1$.

For $f(x)$ to be onto, the codomain $B$ must be equal to the range of $f(x)$ on $[1, \infty)$.

Since $f(x)$ is strictly increasing on $[1, \infty)$, its minimum value occurs at $x = 1$:

$f(1) = 2(1)^3 - 3(1)^2 + 6 = 5$

As $x \rightarrow \infty$, $f(x) \rightarrow \infty$.

Thus, the range is $[5, \infty)$, which means $B = [5, \infty)$.

Answer: $a = 1, B = [5, \infty)$