1. For the quadratic equation $ax^2+bx+c=0,$ with $a,b,c\in\{1,2,3,4,5,6\}$, the total number of possible equations is $6\times 6\times 6=216.$

2. The equation has one real root bigger than the other means it has two distinct real roots. For a quadratic, this happens when the discriminant is positive: $D=b^2-4ac>0.$

3. So we must count the number of ordered triples $(a,b,c)$ such that $b^2>4ac.$

4. Since $b\in\{1,2,3,4,5,6\}$, we check each value of $b$.

Case 1: $b=1$

$b^2=1$ Need: $4ac<1$ Impossible since $a,c\ge 1\Rightarrow 4ac\ge 4$.

Number of favorable triples: $0$.

Case 2: $b=2$

$b^2=4$ Need: $4ac<4 \Rightarrow ac<1$ Impossible.

Number of favorable triples: $0$.

Case 3: $b=3$

$b^2=9$ Need: $4ac<9 \Rightarrow ac<\frac{9}{4}=2.25$ So possible integer products are: $ac=1,2.$ From $a,c\in\{1,2,3,4,5,6\}$, ordered pairs are:

• $(1,1)$
• $(1,2)$
• $(2,1)$

Count = $3$.

Case 4: $b=4$

$b^2=16$ Need: $4ac<16 \Rightarrow ac<4$ So possible integer products are: $ac=1,2,3.$ Ordered pairs:

• $ac=1$: $(1,1)$
• $ac=2$: $(1,2),(2,1)$
• $ac=3$: $(1,3),(3,1)$

Count = $1+2+2=5$.

Case 5: $b=5$

$b^2=25$ Need: $4ac<25 \Rightarrow ac<6.25$ So possible integer products are: $ac=1,2,3,4,5,6.$ Ordered pairs:

• $ac=1$: $(1,1)$
• $ac=2$: $(1,2),(2,1)$
• $ac=3$: $(1,3),(3,1)$
• $ac=4$: $(1,4),(2,2),(4,1)$
• $ac=5$: $(1,5),(5,1)$
• $ac=6$: $(1,6),(2,3),(3,2),(6,1)$

Count = $1+2+2+3+2+4=14$.

Case 6: $b=6$

$b^2=36$ Need: $4ac<36 \Rightarrow ac<9$ So possible integer products are: $ac=1,2,3,4,5,6,8.$ (Note: $ac=7$ is not possible with factors in the set.)

Ordered pairs:

• $ac=1$: $(1,1)$
• $ac=2$: $(1,2),(2,1)$
• $ac=3$: $(1,3),(3,1)$
• $ac=4$: $(1,4),(2,2),(4,1)$
• $ac=5$: $(1,5),(5,1)$
• $ac=6$: $(1,6),(2,3),(3,2),(6,1)$
• $ac=8$: $(2,4),(4,2)$

Count = $1+2+2+3+2+4+2=16$.

5. Total favorable triples: $0+0+3+5+14+16=38.$

Thus, $p=\frac{38}{216}.$ Therefore, $216p=38.$

6. So the correct option is $\boxed{38}.$

This matches Option A.