jmc

Since $-2-3i$ is a root $$a(-2-3i{)}^3+3(-2-3i{)}^2+b(-2-3i)-65=0.$$Expanding, we get $$(-80+46a-2b)+(36-9a-3b)i=0.$$Then $-80+46a-2b=0$ and $36-9a-3b=0.$ Solving, we find $a=2$ and $b=6.$

The cubic polynomial is then $2x^3+3x^2+6x-65=0,$ which factors as $(2x-5)(x^2+4x+13)=0.$ Therefore, the real root is $\boxed{\frac{5}{2}}.$