We can describe the given language as a set of all strings over alphabet having EQUAL and EVEN numbers of 0's,2's and 4's.
Let DFA for the given language has K states.
Take X= 02k22k42k which belongs to L.(as 2k will be the even number and the number of 0's=number of 2's =number of 4's).
Now however you divide the string into u,v,w "pumping part" will always contain all 0's only as K<2K.
So after dividing X into u,v and w, if we pump 'v' part for, 'm' times we will get AT LEAST 'm' additional 0's.
but, 02k+m22k42k ∉ L as the number of 0's > number of 2's and 4's.
Hence language is not regular.
According to me, L is a CSL.
@arjun sir please check whether the proof is correct or not.