$a\lambda^3 – b \lambda^2 + c \lambda +d = \lambda^3 – 4 \lambda^2 + x \lambda +30$
(Using $x$ in place of $a$ on question to preserve the convention)
$\implies a = 1, b = -4, c = x, d = 30$
Sum of eigen values $= (-b/a)= 4.$
So, $x_1 + x_2 + x_3 = 4$
Given $x_1 = 2$
So, $ x_2 + x_3 = 4 - 2 = 2 \quad \to (1)$
Product of eigen values $= (-d/a) = -30.$
So, $x_1 . x_2 . x_3 = -30$
$\implies x_2 . x_3 = -15 \quad \to (2)$
Solving $(1)$ and $(2)$, we get
$x_2 = -3 , x_3 = 5$
Hence, the absolute value of largest of 3 eigen values = 5.