The total cost should be:
(cost for driver team) + (cost of fuel) + (cost to keep the truck on the road)
The cost of the driver team is 27h, as you have written. The cost to keep the truck on the road is 15h. The "over and above the cost of fuel" doesn't mean that you should divide this by the cost of fuel per gallon, it means "in addition to fuel cost". So you need to figure out the cost of fuel, which is where you will use the mpg.
This exercise can be managed by using the EOQ-formula.
In your case the total costs are
$TC=\frac{K \cdot D}{Q}+\frac{Q\cdot h}{2}=\frac{320 \cdot 180,000}{Q}+\frac{Q\cdot 20}{2}$
$P$ = store costs per year
$Q$ = produce quantity
$D$ = annual demand quantity
$K$ = Set up production cost
$h$=storage cost per unit
The derivative w.r.t Q is
$\frac{\partial TC}{\partial Q}=-\frac{K \cdot D}{Q^2}+\frac{ h}{2}=0$
Solving for Q
$\frac{K \cdot D\cdot 2}{h}=Q^2$
$Q^*=\sqrt{\frac{K \cdot D\cdot 2}{h}}$
After you have calculated the optimal produce quantity the number of cost minimizing set ups is $\frac{D}{Q^*}$
Best Answer
Assumptions
The question is lacking in some specifics so here are my assumptions:
The mice are used up at an equal rate over the year. As such the food requirements decreases (or there would be no real need for calculus). For example buying 600 mice at the start of a year would end up only needing feed for 300 as the mice get used up over the year. So each mice effectively only eats \$2 worth of food.
Breakdown
The problem is now about how often to order the mice vs how long the need storage/feeding. If we buy infrequently we have bigger feeding costs but low service fee. If we buy frequently we have low feeding costs buy high service fee.
Solution
Let us order mice $n$ times per year. The service fee is therefore $12n$. We will be keeping $\frac{600}{n}$ mice for $\frac{1}{n}$ of a year. From above each mouse only eats the $\frac{\$2}{n}$ of food for this period. This occurs $n$ times so the food cost is:
$$n\times\frac{600}{n}\times\frac{2}{n}=\frac{1200}{n}$$
So the total cost is: $$C=12n+1200n^{-1}$$
Calculus can be used to find the minimum.
$$\frac{dC}{dn}=12-1200n^{-2}$$
The minimum will occur when $\frac{dC}{dn}=0$
$$0=12-1200n^{-2}$$
$$1200n^{-2}=12$$
$$100=n^2$$
$$n=10$$
As ${\frac{dC}{dn}}_{n=9}<0$ and ${\frac{dC}{dn}}_{n=11}>0$ we can see that $n=10$ is a minimum and not a maximum.
So 60 mice should be bought ten times per year.