Given:
- Volume \( V = 300 \, {litre} = 0.3 \, {m}^3 \)
- Mass of oxygen \( m = 3 \, {kg} \)
- Molar mass of oxygen \( M = 32 \, {kg/kmol} \)
- Temperature \( T = 25^\circ {C} = 298.15 \, {K} \)
First, calculate the number of moles \( n \) of oxygen:
\[
n = \frac{m}{M} = \frac{3}{32} = 0.09375 \, {kmol}
\]
Now, using the ideal gas law:
\[
PV = nRT
\]
Rearrange to find the pressure \( P \):
\[
P = \frac{nRT}{V}
\]
Substitute the values:
\[
P = \frac{0.09375 \times 8.314 \times 298.15}{0.3} = 773.07 \, {kPa}
\]
Thus, the pressure inside the tank is approximately \( 773.07 \, {kPa} \), which lies between 773 and 776 kPa.