= Solution
The intact source figure gives <orbital period> $P=6.0\,\mathrm{yr}$, stellar <radial velocity> semiamplitude $K=30.0\,\mathrm{m\,s^{-1}}$, and <orbital eccentricity> $e=0$. A planetary <exoplanet transit> implies a nearly edge-on orbit, so $\sin i\simeq1$. Adopt a <solar mass> for the <Solar-type star> and initially neglect the companion's <mass> relative to it.
For a circular solution of the <two-body problem>, the star's distance from the <centre of mass> is $a_*=aM_p/(M_*+M_p)$, and the measured semiamplitude is $K=2\pi a_*\sin i/P$. <Kepler's third law> gives
$$
a^3=\frac{G(M_*+M_p)P^2}{4\pi^2},\qquad
\frac{M_p^3\sin^3 i}{(M_*+M_p)^2}=\frac{PK^3}{2\pi G}.
$$
Hence
$$
\boxed{a\simeq\left(\frac{GM_\odot P^2}{4\pi^2}\right)^{1/3}=3.30\,\mathrm{AU},\qquad
M_p\simeq K\left(\frac{P}{2\pi G}\right)^{1/3}M_\odot^{2/3}=3.64\times10^{27}\,\mathrm{kg}\simeq1.92M_J.}
$$
The inferred <binary mass ratio> is only $0.00183$, validating the approximation. Since $e=0$, the star–planet separation is constant and equals the orbital <semimajor axis>. Without the <exoplanet transit>, <Doppler spectroscopy> would generally give only a minimum companion <mass> through $M_p\sin i$.
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