There must be spontaneous fission of uranium taking place in the Sun. But the long half-life of this process, combined with the very low abundance of uranium means that fission is a totally negligible fraction ($\sim 10^{-26}$) of the average solar output. The fission process is spontaneous rather than induced by neutrons, because there is no significant source of free neutrons in the Sun.
The energy produced by radioactive decay of uranium is far larger, because the lifetimes for radioactive decay are shorter, although the total output is still negligible - a fraction of about $1.4\times 10^{-10}$ of the solar output, averaged over its lifetime.
Details
Spontaneous Fission
There will of course be some fission taking place in the Sun. That fission would have to be spontaneous since there is no plentiful supply of free neutrons inside the Sun. The spontaneous fission process itself is so slow, and the fissile material so sparse that any kind of "chain reaction" is not possible.
The Uranium does not get concentrated significantly towards the core. That is because the Sun is sufficiently well mixed and turbulent that any diffusion processes are rather slow. Asplund et al. (2020) estimate the primordial abundance fraction of uranium, and accounts both for its half-life to radiactive decay and for a modest amount (only 0.06 dex) of concentration towards the centre over its lifetime. They report initial fractional abundances (by number, with respect to hydrogen) of $10^{-12.65}$ and $10^{-12.16}$ for $^{235}U$ and $^{238}U$ respectively.
The primordial mass fraction of hydrogen is estimated to be 71%. Thus if we take 1 kg of the primordial solar material, then it contains about $4.25\times 10^{26}$ hydrogen atoms, $9.5\times 10^{13}$ $^{235}$U atoms and $2.9\times10^{14}$ $^{238}$U atoms.
The spontaneous fission decay constants, $\lambda$, of $^{235}$U and $^{238}$U can be calculated as $2.0\times10^{-19}$ year$^{-1}$ and $8.3\times 10^{-17}$ yr$^{-1}$ respectively.
The number of spontaneous fissions that have taken place over the lifetime of the Sun will be
$$N_0[1 - \exp(-\lambda\ \tau)] \simeq N_0\lambda\ \tau\ ,$$
(assuming $\lambda\ \tau \ll 1$), where $N_0$ is the initial number of nuclei and $\tau$ is the age of the Sun ($\tau \simeq 4.57\times 10^{9}$ years). This is an upper limit because the uranium nuclei will be decaying other than by fission on shorter lifetimes. Each spontaneous fission ultimately releases about 200 MeV of energy.
Taking the mass of the Sun as $2\times 10^{30}$ kg, then the total number of decays is $3.8\times 10^{25}$ and $4.8\times 10^{28}$ for $^{235}$U and $^{238}$U respectively (i.e. the contribution from the light isotope is negligible because of its lower abundance and slower spontaneous fission rate). This number should probably be reduced by a small factor because the $^{238}$U has a radioactive decay half-life roughly equal to the age of the Sun.
The total energy released by U fission is therefore about $10^{31}$ MeV or $1.5\times 10^{18}$ J over the lifetime of the Sun at an average power of $\sim 10$ Watts.
This is roughly the amount of energy that the Sun releases in 4 billionths of a second!
Radioactive decay
Far more energy is generated from the radioactive decay of uranium into lead isotopes. The rate-determining step here is the half-life of the initial decays of the uranium isotopes ($\lambda = 9.8\times10^{-10}$ year$^{-1}$, and $1.55\times 10^{-10}$ year$^{-1}$ for $^{235}$U and $^{238}U$ respectively.)
The number of decays is given by the equation above (the approximation cannot be used here), thus 99% of the $^{235}$U and about 50% of the $^{238}$U has decayed.
Each decay chain (to lead) releases about 100 MeV of energy. Thus the total energy released by both isotopes is $(0.99\times 9.5\times 10^{13} + 0.5\times 2.9\times 10^{14})\times 2\times 10^{30} \times 100$ MeV $= 7.6\times 10^{33}$ J - an average of $5\times 10^{16}$ W over the solar lifetime, or about $1.4\times 10^{-10}$ of its total output, with the fraction declining from an initially higher value to a value lower than this today (as all of the $^{235}$U has decayed).
