The function you present has no minimum value due to the fact that it is cubic with no other terms. You can use calculus as follows to prove this:
C(v) = 160 + .01 * v^3
C’(v) = .03 * v^2
0 = .03 * v^2
v^2 = 0 / .03 = 0
v = ±0 = 0
The function C'(v) is the derivative of the cost function. By setting this to zero, you are solving for the extrema of the function C(v). Extrema include minimums (what you are looking for), maximums, and points of inflection. In this case, there is a point of inflection at velocity = 0 km/hour. You can check this by either graphing the function or checking the second derivative to figure out which type of extrema the point is. You should notice however that functions of the form v^3 always have a point of inflection at v=0 with no local maximums or minimums to be found.
To answer your question however based on the function, because the cost per hour gets exponentially larger the faster the ship travels, the slower the ship goes, the more cost efficient the journey will be. In fact, as the ships speed approaches zero, the cost of running the ship approaches $160 per hour. This however is assuming the cost is independent of the distance to be traveled...
...however, from reading part (a) and part (b) of your question, I assume you can use other variables to express your answer, and in this case, the distance of the journey is crucial in solving your problem (distance will be 'd'):
C(v) = 160 + .01 * v^3
t(v) = d / v
The second function there is the length of the journey given d distance and v velocity. By multiplying these functions together, you get the total cost of a journey in terms a constant (given) distance and an unknown velocity. Taking the derivative of this function then allows you to solve for the minimum cost of that journey:
C(v)*t(v) = (160 + .01 * v^3) * (d / v)
cost(v) = 160*d*v^-1 + .01*d*v^2
Knowing the distance to be a given constant you can solve for a minimum cost with velocity 'v' using the above equation. As part (b) to your question alludes, the faster the ship goes, the more cost effective it will be up until a point. In the case of 16 km/hr being the maximum speed of the ship, the following would be the minimum cost of the journey in terms of 'd' distance:
cost(v) = 160*d*v^-1 + .01*d*v^2
cost(v) = 160*d*(16)^-1 + .01*d*(16)^2
cost(v) = 10*d + 2.56*d
cost(v) = 12.56*d
In this case, the cost of running the ship is $12.56 per hour if the ship goes at its maximum speed throughout the journey. Ultimately though, the pitfall to this kind of question is over-analyzing the question using math. Compuchip's response is 'correct' in a sense of the word, but doesn't truly answer the question being asked, and the question is fairly vague and sounds like it's missing a part considering it asks about a 'journey' in part (a) but doesn't given any facts about the journey (ie distance).
