Step 1: Get the actual travel speed in metres per minute.
Slip cuts the theoretical speed of 8 km/h by 8%, giving an actual speed $V_a = 8 \times 0.92 = 7.36$ km/h.
Converting, $V_a = \dfrac{7.36 \times 1000}{60} = 122.7$ m/min.
Step 2: Get the swath width covered by the boom.
Half the spray angle is $55^{\circ}$, so one nozzle's footprint at 0.60 m height spans $w = 2(0.60)\tan 55^{\circ} = 1.714$ m.
Since adjacent footprints must overlap by 30%, nozzles sit $0.70 \times 1.714 = 1.200$ m apart, and 12 of them give a swath of $12 \times 1.200 = 14.40$ m.
Step 3: Get the ground area covered per minute.
Raw coverage rate $= W \times V_a = 14.40 \times 122.7 = 1766.9\ m^2/min$.
Only 75% of this counts as useful field work because of turns and overlaps in field efficiency, so the effective rate is $1766.9 \times 0.75 = 1325.2\ m^2/min$.
Step 4: Convert the field size and divide.
$25$ ha $= 250000\ m^2$.
\[ t = \frac{250000}{1325.2} = 188.7\ min \]
Final Answer:
Rounding to the nearest minute, spraying the field takes
\[ \boxed{189\ minutes} \]