Let's solve the given triple integral one step at a time.

The integral is defined as

$$\int\limits_{ - 3}^3 {\int\limits_{ - 2}^2 {\int\limits_{ - 1}^1 {(4{x^2}y - {z^3})dz\,dy\,dx} } }$$

First, we'll integrate with respect to $z$, treating $x$ and $y$ as constants:

$$\int\limits_{ - 1}^1 {4x^2y - z^3} dz$$

We integrate the function term by term:

$$\int\limits_{ - 1}^1 {4x^2y\,dz} - \int\limits_{ - 1}^1 {z^3\,dz}$$

The first term $4x^2y$ is constant with respect to $z$, so its integral is simply $4x^2y$ times the width of the interval, which is $z$ evaluated from $-1$ to $1$. The second term $\int{z^3\,dz}$ is an odd function; since we're integrating over a symmetric interval around zero, its integral will cancel out.

$$4x^2y \cdot \left( {1 - ( - 1)} \right) - \left[ {\frac{{z^4}}{4}} \right]_{ - 1}^1$$

$$= 8x^2y - \left( {\frac{{1^4}}{4} - \frac{{(-1)^4}}{4}} \right)$$

$$= 8x^2y - \left( {0} \right)$$

$$= 8x^2y$$

Now we have a simpler double integral:

$$\int\limits_{ - 3}^3 {\int\limits_{ - 2}^2 {8{x^2}y\,dy\,dx} }$$

Next, we integrate with respect to $y$:

$$8x^2 \int\limits_{ - 2}^2 {y\,dy}$$

This integral is also of an odd function (\(y\) is odd), and given that the interval is symmetric around zero, the integral evaluates to 0:

$$8x^2 \cdot \left[ \frac{y^2}{2} \right]_{ - 2}^2$$

$$= 8x^2 \cdot \left( \frac{2^2}{2} - \frac{(-2)^2}{2} \right)$$

$$= 8x^2 \cdot (0)$$

$$= 0$$

The value we've just found is the integrand of the last integral. Since the integrand is 0, the last integral with respect to $x$ will also be 0, irrespective of the limits:

$$\int\limits_{ - 3}^3 {0\,dx}$$

The value of the whole triple integral is:

$$0$$

Thus, when rounded off to the nearest integer, the value of the definite integral is 0.