 Solution

The common one-per-cent exoplanet transit depth gives $R_p=0.1R_{\star }$. Apply the supplied main-sequence mass-radius relation, $R_{\star }/R_{\odot }=M_{\star }/M_{\odot }$, and Kepler's third law, neglecting the planet mass. For $P=6$ days,The companions are approximately Jupiter-sized and Earth-sized respectively; these are sizes, not independently determined masses. The duration uses the small-planet chord approximation. Counting the finite planet radius would increase first-to-fourth-contact durations by about ten per cent for central transits.

For equal distance, passband, cadence and noise dominated by stellar photon counting, the supplied mass-luminosity relation makes the solar-mass host about $10^3$ times more luminous, with a transit about $4.64$ times longer. At equal fractional exoplanet transit depth, signal-to-noise per event scales as $\delta \sqrt{{\dot{N}}_{\gamma }{t}_{\mathrm{tr}}}$, favouring the solar-mass host by about $\sqrt{1000\times 4.64}\simeq 68$. Equal apparent brightness would remove the luminosity advantage but retain the duration advantage. Without distances and noise properties the relative detectability is conditional. Transit probability is also larger for the solar-mass case, about $7.2\%$ versus $1.55\%$.