Convert 251_m to base 10.

To convert a number from base $m$ to base 10, we multiply each digit by $m$ raised to the power of its position, starting from 0 for the rightmost digit.

$$251_m = 2 \times m^2 + 5 \times m^1 + 1 \times m^0$$ $$251_m = 2m^2 + 5m + 1$$

We are given that $251_m = 100_{\text{ten}}$.
So, we set the base 10 equivalent equal to 100:

$$2m^2 + 5m + 1 = 100$$

Subtract 100 from both sides:

$$2m^2 + 5m + 1 - 100 = 0$$ $$2m^2 + 5m - 99 = 0$$

We can solve this quadratic equation using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=2$, $b=5$, and $c=-99$.

$$m = \frac{-5 \pm \sqrt{5^2 - 4(2)(-99)}}{2(2)}$$ $$m = \frac{-5 \pm \sqrt{25 + 792}}{4}$$ $$m = \frac{-5 \pm \sqrt{817}}{4}$$

Calculate the square root of 817: $\sqrt{817} \approx 28.583$.

$$m = \frac{-5 \pm 28.583}{4}$$

This gives two possible values for $m$:

$$m_1 = \frac{-5 + 28.583}{4} = \frac{23.583}{4} \approx 5.896$$ $$m_2 = \frac{-5 - 28.583}{4} = \frac{-33.583}{4} \approx -8.396$$

Alternatively, we can try to factor the quadratic equation. We look for two numbers that multiply to $2 \times (-99) = -198$ and add to 5. These numbers are 18 and -11.
We rewrite the middle term:

$$2m^2 + 18m - 11m - 99 = 0$$

Factor by grouping:

$$2m(m + 9) - 11(m + 9) = 0$$ $$(2m - 11)(m + 9) = 0$$

This gives two possible values for $m$:

$$2m - 11 = 0 \implies 2m = 11 \implies m = \frac{11}{2} = 5.5$$ $$m + 9 = 0 \implies m = -9$$

A number base must be a positive integer and greater than the largest digit in the number. The largest digit in $251_m$ is 5.
Therefore, $m$ must be an integer and $m > 5$.
From our solutions, $m = 5.5$ is not an integer, and $m = -9$ is not positive.
There seems to be an issue with the problem statement or the expected answer if $m$ must be an integer. However, if we consider $m$ can be non-integer, $m=5.5$ is the only positive solution. But bases are typically integers.

Let's re-check the calculation.
$2m^2 + 5m - 99 = 0$
If $m=6$: $2(6^2) + 5(6) - 99 = 2(36) + 30 - 99 = 72 + 30 - 99 = 102 - 99 = 3 \neq 0$.
If $m=7$: $2(7^2) + 5(7) - 99 = 2(49) + 35 - 99 = 98 + 35 - 99 = 133 - 99 = 34 \neq 0$.

Given the context of typical base problems, $m$ is usually an integer. If $m$ must be an integer, then there is no integer solution for $m$ that satisfies the equation $2m^2 + 5m - 99 = 0$ and $m > 5$.
However, if we strictly follow the calculation, the value of $m$ is 5.5.

Assuming the question implies an integer base, there might be a typo in the question. If we must provide an answer based on the given numbers, $m=5.5$ is the positive solution.

Let's assume the question expects an integer base. If we consider the possibility of a slight error in the problem statement, and if the number was $241_m = 100_{\text{ten}}$, then $2m^2 + 4m + 1 = 100 \implies 2m^2 + 4m - 99 = 0$. This also doesn't yield an integer.

Let's stick to the calculation.
The value of $m$ is $\boxed{5.5}$.