How many seconds are a meter in the 4th dimension? The title really does say it all. 
I want to know how many seconds elapse between a meter in the 4th dimension?
I don't know if this question is naive, or my understand of the dimension is off. 
I'm thinking of a 4d cube with 3d slices; the breadth of the 4d cube which lies along the 4th dimension is time, right?
Does this breadth have no length?
 A: Everybody so far interpreted the question as "how long does light take to move 1m". Let me interpret it in a different way.

How many seconds are a meter in the 4th dimension?

They are not. 
The 3 dimensions of space are different than the 1 dimension of time. 
Interpretation of the question
Consider this experiment (no rockets, spaceships or black holes involved, simply you sitting in front of your computer on earth):
Name our dimensions X, Y, Z and T. Lay a piece of standard paper on your desk. Draw a horizontal line on it. It will have a certain extent in X (let's call it 10cm), in Y (0.1cm - the width of the stroke of your pen) and Z (0.000001cm or however thick the layer of paint is). Label your desk with a "X" and "Y" axis, and if you wish a little "Z" along one of its legs. Do not label the paper with a coordinate system, we don't need it.
Now turn the paper 90° around the Z axis; it will now have 0.1cm length in X and 10cm in Y (remember, the axes are on your desk, not your paper - the desk is your coordinate system / reference system for this experiment). You did not change the line or the piece of paper at all. You merely rotated it around an axis in another spatial dimension. You can do the same with any kind of rotation, and nobody will be able to tell a difference, as long as you don't fold the paper. If you take very stiff paper and somehow glue it to your desk in a standing position, you can also add a Z length by rotating it "upwards".
Whatever kind of rotation you do, you will change the extent of the line in any of the dimensions (related to your desk, not the paper itself).
All of this was just a verbose setup to ask your question another way: 

How many cm are a X-cm in the Y dimension?

The answer is: "the same". After rotation as above, if the line was 10cm in X before, it will be 10cm in Y. That is, the space dimensions (at least the 3 we know of for sure) have the same "spacing" if you so wish, at least in our everyday settings.
(Bear with me, all of this is totally trivial and I wrote it just so you can see how I understood your question.)
About time...
What length does the line have in T? Well, you set a timer when you draw it, and after playing around with it for 10s, you take an eraser and erase the line. This gives you a "length" of 10s in the T dimension.
Now: You can not do the "rotation trick" with T. You cannot turn the line so it is 0cm in X, Y, Z and some different duration in T - you cannot transform space into time. Not theoretically, not practically. No matter what you do (while staying at your desk and not in a rocket ship) will change the line with respect to T in any way whatsoever, as long as you don't go in there with a pen or an eraser, or you tear up the paper. That's also why we would usually call its extent in T something like "duration", not "length". 
Space is space, and time is time, even in spacetime.
Addendum
To give you an idea past this "naive" stuff, check out https://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals_in_flat_space for what happens if you put time and space together in a formula... the inherent difference between space and time dimensions gives rise to different types of spacetime-distances with fundamentally different properties (heck, one is even negative...).
A: The speed of light is 299792458 meters per second, so it would take $1/299792458$ seconds for the photon to travel one meter.
Best of luck.
A: Light travels about $3\times 10^8$ meters in a second, so a meter is about $1/(3\times 10^8)$ seconds.
A: A meter in the fourth dimension is $\frac{i}{c}$ seconds. A four dimensional Euclidean space with metric:
$$ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2$$ 
can be obtained from the Lorentzian metric:
$$ds^2 = -c^2 dt^2 + dx_1^2 + dx_2^2 + dx_3^2$$
by putting $dt = \frac{i}{c} dx_4$
