Hi Sandy.
I'm going to assume that equation you gave is C = 0.136 + 0.001V^{3}.
So let's say before speed reduction, C_{1 } = 0.136 + 0.001V_{1 }^{3} and after speed reduction, C_{2 } = 0.136 + 0.001V_{2 }^{3}.
If we subtract the second equation from the first, we get C_{1 }  C_{2} = 0.001(V_{1 }^{3}  V_{2 }^{3}). You know that consumption dropped by 42 tonnes per day, so that is 42/24 = 1.75 tonnes/hour. That can replace C_{1 }  C_{2} in the left side of the equation. You also know that the speed was reduced by 20%. That means 80% of V_{1} is V_{2}. Now you can substitute the last equation with the replacements: 1.75 = 0.001 [V_{1 }^{3}  (0.8 V_{1 })^{3 }]
So now you can solve for the original speed V_{1}. Use V_{1} to calculate the reduced speed V_{2}, then use these to calculate the original and new consumption rates, C_{1} and C_{2}. I assume that part (a) is a percentage question, since you are given the actual reduction as 42 tonnes per day.
Once you know V_{2}, you can calculate the time to travel 800 nautical miles, then apply that to the fuel consumption rate C_{2}. Add that to the 50 tonnes you finish with and you have your answer to (b).
Part (c) is just a variation on (b) using the original speed and consumption figures, and once you've done the work for (b) and (c), you have the times required for each speed so the difference is the answer to (d).
Hope this helps!
Stephen La Rocque.
