Find. The hydraulic detention (retention) time, $\theta$, and the sludge (solids) retention time, $\theta_c$ (SRT), of the reactor.
Approach. The hydraulic detention time follows directly from reactor volume and flow rate. Since biomass in the supernatant/effluent is stated to be negligible, essentially all the biomass in the system leaves via wasting, so the SRT is the total biomass mass held in the reactor divided by the daily wastage (mass) rate.
Total biomass held in the reactor. Using the (settled/reactor) biomass concentration over the full reactor volume:
$$M_X = X\times V = 15\ \text{g/L}\times200\ \text{m}^3 = 15\ \text{kg/m}^3\times200\ \text{m}^3 = 3000\ \text{kg}$$
Sludge (solids) retention time. Since effluent (supernatant) biomass is ignored, the SRT is total biomass mass over the daily wastage rate, $\theta_c = M_X/P_x$:
$$\theta_c = \frac{3000\ \text{kg}}{300\ \text{kg/d}} = \boxed{10\ \text{days}}$$
Final results
Quantity
Value
Hydraulic detention time, $\theta$
0.5 d (12 hr)
Biomass held in reactor, $M_X$
3000 kg
Sludge retention time, $\theta_c$
10 days
Check: assumes the "settled biomass concentration" of 15 g/L represents the reactor's own (mixed-liquor) biomass concentration used to size the total mass held in the tank — consistent with a recycled (activated-sludge-type) reactor where the wasted sludge is drawn at essentially the same concentration as the settled solids in the system, and with the question's own instruction to ignore effluent/supernatant biomass entirely.
