For hydrogen like atom, find velocity from potential energy PE of system is given by $$\mathrm{U(r) = \frac{-Ke²}{r^3}}$$ assume bohr model to be valid Find velocity


For this I used $\mathrm{\frac{-PE}{2} =KE}$ and then used $$\frac{mv^2}{2} =\frac{Ke^2}{2r^3}$$ and $$mvr=\frac{nh}{2\pi}$$

However, I am getting the answer as $$v=\frac{n^3h^3}{16Ke^2\pi^3m^2}$$  but the answer given has a factor of 24 rather than 16. I don't know where I'm going wrong.
The fault in your solution is the part where you assume  $\mathrm{\frac{-PE}{2} =KE}$ this is true only if the force follows inverse square law.

In this particular case we need to find the force first by using

$$F =\frac{-dU}{dr}$$

Which gives us

$$F =\frac{-3Ke^2}{r^4}$$

Now this force is central and provides centripetal force

$$\frac{mv^2}{r} =\frac{3Ke^2}{r^4}$$ 

This along with 
$$mvr=\frac{nh}{2\pi}$$

gives us
$$v=\frac{n^3h^3}{24Ke^2\pi^3m^2}$$