Here, $Y$ is uniformly distributed in $(1,6)$
$\therefore \: f(Y) = \begin{cases} \frac{1}{b-a} & 1<Y<6 \\ 0, & \text{otherwise} \end{cases}$
$\implies f(Y)=\frac{1}{b-a} = \frac{1}{6-1} = \frac{1}{5}$
Now given Polynomial,
$3x^2+6xY+(3Y+6)$
For real roots $b^2-4ac \geq 0$
$\implies (6Y)^2 - 4 \times 3 \times (3Y+6) \geq 0$
$\implies Y^2-2Y+Y-2 \geq 0$
$\implies (Y-2)(Y+1) \geq 0$
$\implies Y \geq 2, \: Y \geq -1$
So, we need area of the region in the above graph, for $(2,6)$
Area $= B \times H$
$=(6-2) \times \frac{1}{5}$
$= \frac{4}{5} = 0.8$