The characteristic polynomial $\lambda^3-4\lambda^2+a\lambda+30$ is three dimensional. So there will be 3 roots of this polynomial equation i.e. $\lambda^3-4\lambda^2+a\lambda+30=0 \tag{i}$
Putting $\lambda=2$ (as given in the question) to no$\mathrm{(i)}$, we get
$2^3-4\times2^2+2a+30=0\Rightarrow a=-11$.
So the characteristic polynomial equation is now $\lambda^3-4\lambda^2-11\lambda+30=0$. Since one of its roots is $2$, hence $(\lambda -2)$ will divide the polynomial.
So
$\begin{align}\lambda^3-4\lambda^2-11\lambda+30&=0\\\Rightarrow \lambda^2(\lambda-2)-2\lambda^2-11\lambda+30&=0\\\Rightarrow \lambda^2(\lambda-2)-2\lambda(\lambda-2)-15\lambda+30&=0\\\Rightarrow \lambda^2(\lambda-2)-2\lambda(\lambda-2)-15(\lambda-2)&=0\\\Rightarrow (\lambda-2)(\lambda^2-2\lambda-15)&=0\\ \therefore \lambda^2-2\lambda-15&=0\\\Rightarrow \lambda &= \frac{2\pm\sqrt{2^2+4\times15}}{2}\\ &=1\pm4 \\ &=5,-3 \end{align}$
$\therefore$ The Eigenvalues $\lambda=-3,2,5$.
So the largest absolute of $\lambda$ i.e. the maximum absolute Eigenvalue is $5$.