A set of four consecutive multiple of four, and its average is 30. If another series which is five consecutive even number series which lowest number is equal to second lowest multiple of these four consecutive multiples of four. Find the highest number of five consecutive even number series?

Let us find the four consecutive multiples of 4 first.
Let the four consecutive multiples of 4 be $x$, $x+4$, $x+8$, and $x+12$.

The average of these four numbers is given as 30.
Therefore, we can write the equation:
$\frac{x+(x+4)+(x+8)+(x+12)}{4}=30$

Summing the terms in the numerator:
$4x+24=120$
$4x=96$
$x=24$

So, the four consecutive multiples of 4 are:
First multiple: $x=24$
Second lowest multiple: $x+4=28$
Third multiple: $x+8=32$
Fourth multiple: $x+12=36$

Now, we are given a second series consisting of five consecutive even numbers.
The lowest number of this new series is equal to the second lowest multiple of the first series, which is 28.

Let the five consecutive even numbers be:
First (lowest): 28
Second: 30
Third: 32
Fourth: 34
Fifth (highest): 36

Thus, the highest number of this five consecutive even number series is 36.