The quadratic equation $x^2+bx+c=0$ has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of $b^2+c$?

Given that the roots of the quadratic equation $x^2+bx+c=0$ are $4a$ and $3a$, where $a$ is an integer.

Using the relations between roots and coefficients:
Sum of roots: $4a+3a=-b\Rightarrow b=-7a$
Product of roots: $\left(4a\right)\left(3a\right)=c\Rightarrow c=12a^2$

We need to find a possible value of $b^2+c$:
$b^2+c={(-7a)}^2+12a^2=49a^2+12a^2=61a^2$

Since $a$ is an integer, $b^2+c$ must be a multiple of 61. Let us check the options:
1. $3721/61=61$, which means $a^2=61$. Since 61 is not a perfect square, $a$ is not an integer.
2. $549/61=9$, which means $a^2=9\Rightarrow a=\pm 3$. Since 3 is an integer, this is a possible value.
3. $427/61=7$, which means $a^2=7$. Since 7 is not a perfect square, $a$ is not an integer.
4. $361/61\approx 5.92$, which is not an integer.

Therefore, the only possible value is 549.