Distribution coeffecient without concentrations
Molar concentration can be expressed via mass $m$ and volume $V$ all right:

$$C_i = \frac{n_i}{V_i} = \frac{m_i}{M_iV_i}$$

Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = \text{const})$ so the distribution coefficient $D$ can be rewritten as such:

$$D = \frac{C_\mathrm{s}}{C_\mathrm{w}} = \frac{m_\mathrm{s}V_\mathrm{w}}{m_\mathrm{w}V_\mathrm{s}}$$

where "s" refers to the solvent phase and "w" to aqueous phase.
At equilibrium

$$m_\mathrm{w} = m_0 - m_\mathrm{s}$$

where $m_0$ is the initial mass of the analyte.
Finally, the distribution coefficient is

$$D = \frac{m_\mathrm{s}V_\mathrm{w}}{(m_0 - m_\mathrm{s})V_\mathrm{s}} = \frac{\pu{0.0376 g}\cdot\pu{150 mL}}{(\pu{0.192 g} - \pu{0.0376 g})\cdot\pu{50 mL}} = 0.731$$