= Solution
Assume the companion is much less massive than the star, the sinusoid represents a circular Keplerian orbit, and the transit implies $\sin i\simeq1$. Ignore dilution, limb darkening and grazing geometry when estimating the radius. With $P=5\times86400\,\mathrm s$,
$$
a\simeq\left(\frac{GM_\odot P^2}{4\pi^2}\right)^{1/3}
=8.56\times10^9\,\mathrm m=0.0572\,\mathrm{AU}.
$$
The stellar reflex <radial-velocity semi-amplitude> gives $K\simeq(m/M_\odot)(2\pi a/P)$, while transit depth gives $\delta\simeq(R_p/R_\odot)^2$. Therefore
$$
\boxed{m\simeq1.12\times10^{27}\,\mathrm{kg}=0.589M_{\rm Jup},\qquad
R_p\simeq9.84\times10^7\,\mathrm m=1.408R_{\rm Jup}.}
$$
The <mean density> relative to the supplied Jupiter values is
$$
\boxed{\frac{\bar\rho_p}{\bar\rho_{\rm Jup}}\simeq\frac{0.589}{1.408^3}=0.211,\qquad
\bar\rho_p\simeq2.80\times10^2\,\mathrm{kg\,m^{-3}}.}
$$
This is a low-density hot gas giant. Strong irradiation can slow contraction or help sustain an inflated envelope; internal heating and a different heavy-element fraction are other possibilities. The inferred low mass also reduces gravitational compression. Mass and radius alone do not determine which inflation mechanism operates.
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