Given:
\( \dot{m}_{fuel}=120\,\text{kg/s} \), \( V_{e}=2300\,\text{m/s} \), \( p_{e}=90\,\text{kPa} \), \( D_{e}=⌀0.3\,\text{m} \).
To calculate: Thrust at sea level.

The schematic diagram of the problem description is shown in Fig. 1.

Assuming the atmospheric pressure at sea level as \( p_{\text{atm}}=1\,\text{atm}=101325\,\text{Pa} \).

Assuming that the oxidizer is mixed with the fuel, \( \dot{m}_{e}=\dot{m}_{fuel}=120\,\text{kg/s} \).

Applying the conservation of momentum on the control-volume around the rocket,

$$ \begin{aligned} \text{Thrust} =&\ \text{rate of momentum exiting }-\text{ rate of momentum entering}\\ &\ +\text{ pressure force at exit }-\text{ pressure force at inlet}\\ \\ \text{Thrust} =&\ \dot{m}_{e}V_{e}-0+\left(p_{e}-p_{\text{atm}}\right)A_{\text{exit}}\\ \\ \text{Thrust} =&\ 120\times2300-0+\left(90\times10^{3}-101325\right)\times\frac{\pi}{4}\times0.3^{2} \end{aligned} $$ $$ \boxed{\text{Thrust at sea level}=275199.483\,\text{N}}\ . $$