Your two basic equations are for gravity and for density:
$$a = GM/r^2 \text{ and } \rho=3M/4\pi r^3$$
And you are given a = 5.0 (roughly half of Earth) and r = 5100000 m (roughly 0.8 of the Earth)
So solving gives a mass of about $1.9\times 10^{24}$ and a density of 3500 kg/m^3. The Earth has a density of about 5100.
Here is where your problems start, as you want a powerful magnetic field, and that needs a liquid iron core. But your planet has a density that is substantially less than that of the Earth. Iron cores are heavy. If you have a heavy iron core, but are still have a density of 3500 you need a lot of lighter stuff, like water to compensate. But you now have a "water world" with 300km deep oceans covering the surface (I haven't done the detailed calculation here)
This is your basic problem. We can turn this around: if we fix $\rho$ at 5000, then this gives a planet with a radius of 3500km
With low gravity, it is going to have a hard time holding on to its atmosphere, even with a magnetic field, but perhaps this can be handwaved away.