Note: The problem I discuss below is not the problem that appeared at this site when I first read the problem statement. Apparently the statement of the problem was changed while I was writing my answer, which I think is still sufficient for the original poster to make use of.
The problem asks for how long it will take an object to travel a certain distance (not given explicitly, but we can find this distance) when traveling at a constant speed. So this is a
distance = (rate)(time)
$d = rt$
problem. We are given that $r = 1.5$ in/sec. Using this information gives us
$d = (2.5)t$
We are asked to find the value of $t$, so it's not likely to be among the given information (except maybe for trick questions). That leaves $d$. Can you find the value of $d$? Draw two horizontal lines to indicate the floor and ceiling, and label the distance between the parallel lines as 6 ft. Now draw a line segment straight up from the floor (i.e. the lower parallel line) and stopping somewhere between the parallel lines, and label its length 30 in. Then $d$ is the distance from the top of the line segment to the upper parallel line. Do you see that this distance is 6 ft $-$ 30 in? (If you have already traveled 30 inches into a trip having a total distance of 6 feet, then the distance that remains to be traveled is 6 ft $-$ 30 in.) Since the units of rate (i.e. speed) are inches/second, $d$ needs to be in inches and $t$ needs to be in seconds.
$d =$ 6 ft $-$ 30 in
$d =$ (6 ft)(12 in/ft) $-$ 30 in
$d =$ 72 in $-$ 30 in
$d = 42$ in
Putting this into $d = (2.5)t$ gives
$42 = (2.5)t$
Dividing both sides by 2.5 gives
$t = 42 \div 2.5$
$t =$ 16.8 sec
If you have to do this without a calculator, note that 2.5 = 5/2, so the computation becomes
$42 \div \frac{5}{2} = \frac{42}{1} \cdot \frac{2}{5} = \frac{84}{5} = \frac{80}{5} + \frac{4}{5} = 16 + 0.8 = 16.8$