Assuming you want to maximize the size of the object in your image, then you should place the object close enough to fill the frame of the camera, but no closer than the minimal focus distance of approximately 10 cm.
Your camera's sensor is a 4:3 aspect ratio, meaning there are 4 horizontal pixels for every 3 vertical pixels (when in the landscape orientation). If your object more than 1.33 times wider than it is tall (i.e., > 4/3 width/height), you should frame it for maximal width. Conversely if the object is more than 75% taller than it is wide (i.e., > 3/4 height/width), you should frame it for maximal height.
A 1/2.43" sensor format is about 6.17 mm wide by 4.55 mm tall. It's confusing, but '1/x"' format is not an actual measurement; it's just a name/number, that doesn't specifically indicate a real world dimension. See: Why is a 1" sensor actually 13.2 × 8.8mm? for an explanation.
The magnification \$M\$ of a subject's size \$h_\mathrm{o}\$, compared to its size in the image sensor \$h_\mathrm{i}\$ is:
$$M = {h_\mathrm{i}\over h_\mathrm{o}}\qquad[1]$$
Now, you probably want to leave some head/bottom room around the object, so give it say, 10%, by multiplying \$M\$ by 0.9.
Another formula for magnification relates the lens's focal length \$f\$ and the subject distance \$d_\mathrm{o}\$ (roughly speaking, from the center of the lens to the subject):
$$M = {f\over f-d_\mathrm{o}}\qquad[2]$$
Setting eqs. [1] and [2] equal to each other (and multiplying the right hand side of eq. [1] by the 0.9 "headroom factor") and solving for \$d_\mathrm{o}\$:
$$\begin{align}
{f\over d_\mathrm{o}-f} &= {0.9\times h_\mathrm{i}\over h_\mathrm{o}} \\
{d_\mathrm{o}-f\over f} &= {h_\mathrm{o}\over 0.9\times h_\mathrm{i}} \\
d_\mathrm{o} &= f\left({h_\mathrm{o}\over 0.9\times h_\mathrm{i}} -1\right) \qquad [3]
\end{align}$$
For example, with your camera's lens of 4.74 mm, and assuming we're maximing the object's height of say, 10 cm (100 mm), in the sensor, then your object should be placed about
$$\begin{align}
d_\mathrm{o} &= 4.74\,[\text{mm}]\left(\frac{100\,[\text{mm}]}{0.9\times 4.55\,[\text{mm}]} - 1\right) \\
&= 4.74\,[\text{mm}] \times 23.4 \\
&= 111\,[\text{mm}]
\end{align}$$
111 millimeters (11.1 cm), or approximately 4.4 inches. Note that this is pretty close to your minimal focus distance of approx. 10 cm (Note: I'm assuming that the stated "depth of field" is misnamed and not really depth of field; rather, I assume it's probably the focus distance range of the lens). Therefore, objects smaller than this example's 10 cm high can't be magnified any larger. However, objects larger than 10 cm can be moved farther away from the lens and maximally magnified according to eq. [3].
