Find the center of mass of the solid $E$ bounded by the planes $x=0, x=4, y=0, y=1, z=0$ and $z=3$. Assume the density is $\rho(x, y, z)=z$.

Step 1 ：The center of mass of a solid in three dimensions is given by the triple integral of the position vector weighted by the density of the solid, divided by the total mass of the solid. The total mass of the solid is given by the triple integral of the density over the volume of the solid.

Step 2 ：The center of mass is given by the formula: \[\bar{x} = \frac{1}{M} \iiint_E x \rho(x, y, z) \, dV\], \[\bar{y} = \frac{1}{M} \iiint_E y \rho(x, y, z) \, dV\], \[\bar{z} = \frac{1}{M} \iiint_E z \rho(x, y, z) \, dV\], where M is the total mass of the solid, given by: \[M = \iiint_E \rho(x, y, z) \, dV\].

Step 3 ：In this case, the density \(\rho(x, y, z) = z\), and the solid E is bounded by the planes x=0, x=4, y=0, y=1, z=0 and z=3. So we need to calculate the triple integrals over the volume of the solid.

Step 4 ：The limits of integration for x are 0 and 4, for y are 0 and 1, and for z are 0 and 3.

Step 5 ：The total mass of the solid, M, is calculated to be 18.

Step 6 ：The coordinates of the center of mass, \(\bar{x}\), \(\bar{y}\), and \(\bar{z}\), are calculated to be 2, 1/2, and 2, respectively.

Step 7 ：Final Answer: The center of mass of the solid E is at the point \((\bar{x}, \bar{y}, \bar{z}) = \boxed{(2, \frac{1}{2}, 2)}\).