It's not a novice question at all; in fact, a lot of people misunderstand the relationship between file size and image resolution.
First, an image's actual size--the amount of memory it will consume--is the number of pixels in the image multiplied by the bit depth. So, for example, a typical 3-megapixel photo with dimensions of 2,048x1,536 has 3,145,728 pixels; if it's a 24-bit photo, that means each pixel needs 24 bits of data to describe it, or 75,497,472 bits. Converting the units makes that 9MB.
For completely uncompressed file formats, such as TIFF (without LZW compression) and BMP, the file size on your hard disk will equal the amount of memory you need to hold the image when you, say, copy it to the clipboard or open it in an image-editing package. For compressed file formats such as JPEG, the size on disk can differ greatly from the file size in memory, because when you load it in memory, it is completely decompressed. If that aforementioned 3-megapixel file takes up 1.2MB on your hard disk, for example, by dividing the uncompressed size (9MB) by the compressed size (1.2) you can see how much it was compressed--in this case, 7.5:1. So, in your case, the 350KB file size tells you that your camera is compressing its 2.1-megapixel photos (which would require approximately 6.2MB RAM to load) by 17.6:1. Since most algorithms can compress a file losslessly--without degrading the image--by about 2:1, then it's a pretty good guess that Olympus is using 8:1 lossy compression in your camera.