Listing gives a much better answer. Here I explain how I would approach the problem. The following two facts spring to my mind:
The in-radius of a triangle is $ r = \frac{2\Delta}{P} $ where $\Delta$ and $P$ are the area and perimeter of the triangle. For a right triangle, it takes the simple form $ r = \frac{ab}{a+b+c}, $ where $a,b,c$ are the sides ($c$ is the hypotenuse).
The side lengths of an integer right triangle are given by the Pythagorean triples. I know that any Pythagorean triple1 can be written as $(2kmn, k(m^2-n^2), k(m^2+n^2))$, where $k$, $m$ and $n$ are positive integers. So the inradius becomes $ \frac{2kmnk(m^2-n^2)}{2kmn+(km^2-kn^2)+(km^2+kn^2)} = \frac{2mkn(m-n)(m+n)}{2m(m+n)} = kn(m-n), $ which is an integer.
1If we want only primitive Pythagorean triples, then in the formula, we should restrict $k$, $m$ and $n$ so that $k=1$ (think of $k$ as the scaling factor multiplying a given primitive solution), $\gcd(m,n)=1$ and $m-n$ is an odd integer. (Exercise: What happens when $m$ and $n$ are both odd, so that $m-n$ is even?)
That rules out 2 options, leaving only $5$ as a possibility. If exactly one of the options is guaranteed to be correct, then I would mark this option and move on ;). Otherwise...
We want to find if there are positive integral solutions to $n(m-n)=5$. This gives us two solutions:
$n = 1$ and $m=6$. The triangle in this case is $(12, 35, 37)$.
$n=5$ and $m=6$. The triangle in this case is $(11, 60, 61)$ (rearranging the sides a bit).
If we want a general solution, every positive integer $r$ can be the inradius of an integer right triangle whose sides are primitive integers. (André shows that the $(3r, 4r, 5r)$-triangle has in-radius $r$, which gives an easy solution to the problem if we remove the primitiveness restriction.) It is easy to see that the equation $ r = n(m-n) $ is satisfied by $n=r$ and $m=r+1$, which all the necessary conditions ($\gcd(m,n)=1$ and $m-n=1$, which is odd). This solution gives the triangle $(2r+1, 2r(r+1), 2r^2+2r+1)$. (Of course, this is not the only possible solution.)