A better estimate might be to use the moment of inertia of a $n=3/2$ (fully convective) polytrope, which will be a good approximation in an object like Jupiter, even when it approaches electron degeneracy in its interior.
You can look this up and it is given by $I = kMR^2$, with $k=0.205$ (almost exactly a factor of two smaller than a uniform sphere because the mass is concentrated towards the centre).
If we use a mass $M$ of $1.898 \times 10^{27}$ kg, the equatorial radius of Jupiter ($R=71500$ km) and the rotation rate determined from its magnetosphere ($P= 9.93$ hours), then the angular momentum ($2\pi I/P$) is $3.496 \times 10^{38}$ kg m$^2$ s$^{-1}$.
If instead, we use the volumetric average radius of $R=69900$ km, this reduces to $3.34\times 10^{38}$ kg m$^2$ s$^{-1}$.
This simple approach neglects the more complicated equation of state of Jupiter, it's non-sphericity and the probable presence of a solid core. Consideration of these, together with constraints provided by the measured even harmonics of the gravitational field (referred to as $J_2$ and $J_4$, led Helled et al. (2011) to suggest that $k=0.264$, with an uncertainty below 1%. Combined with the average radius, this gives $4.30\times 10^{38}$ kg m$^2$ s$^{-1}$.
More recent work by Ni (2018) has used refinements to the gravitational harmonics from Juno measurements, along with a more sophisticated interior model to suggest $k=0.274$ with 0.5% uncertainty.