Quadratic Casimir of Higher dimensional representations

(Quadratic) Casimir elements can be represented in terms of elements of the (universal enveloping algebra of the) associated Lie algebra. For SU($n$) this Lie algebra is $frak{su}(n)$. This Lie algebra can be represented in terms of the generators of the group by

$$[T_a,T_b] = f_{ab}^{;;c}T_c$$

Then a (quadratic) Casimir $C$ is one for which

$$[T_a,C]=0$$

for all generators $T_a$.

It sounds like you know the structure constants $f_{ab}^{;;c}$ of your group of interest, and so if you can decompose your formulae for Casimirs in different representations in terms of the corresponding generators in each representation, you should find that they are the same, and that you can then immediately write down the Casimir for your representation of interest by reusing the same formula with the corresponding generators.

You have not shown your work, so it is hard to tell how you are applying your formulae, and plugging in to get your answers.

The Dynkin indices (a,b,c,d) of an SU(5) irrep Λ give you the dimensionality $$ N(a,b,c,d) =(1+a)(1+b)(1+c)(1+d) times left(1+frac{a+b}{2}right) left(1+frac{b+c}{2}right) left(1+frac{c+d}{2}right) left(1+frac{a+b+c}{3}right)left(1+frac{b+c+d}{3}right) times left(1+frac{a+b+c+d}{4}right), $$ Hence $$ N(1,0,0,0)=5 N(0100)=10 N(2000)=15 N(1001)=24 N(0003)=35 N(0011)=40 N(0101)=45 , ...$$ and so on, while their quadratic Casimirs are, in that order, $$ 24/5;~72/10;~24cdot 7/15; ~~~10~~~;~24cdot 28/35; ~ 24cdot 22/40; 24cdot 24/45 ~~~... $$ the last line, 64/5, being the Casimir of the 45.

You may check by your hook rule that (0101) is indeed the antisymmetric traceless 45, of course. (For SU(N) it would be the $N(N(N-1)/2-1)$-dimensional one.) The Young tableau has a 4-story column and a 2-story one.