1. Initial lens setup

The object is at distance $u_1 = 2.4\,\text{m} = 240\,\text{cm}$ and the sharp image is formed on the film at $v_1 = 12\,\text{cm}.$

Using the lens formula, $\frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1}$ with sign convention, or equivalently in magnitudes for a real object and real image, $\frac{1}{f} = \frac{1}{12} + \frac{1}{240} = \frac{20+1}{240} = \frac{21}{240} = \frac{7}{80}.$ So, $f = \frac{80}{7}\,\text{cm}.$

2. Effect of inserting the glass plate

A glass slab of thickness $t = 1\,\text{cm}, \qquad \mu = 1.50$ is inserted between the lens and the film.

For normal incidence, insertion of a slab shifts the image away from the lens by $\Delta = t\left(1 - \frac{1}{\mu}\right).$ Thus, $\Delta = 1\left(1 - \frac{1}{1.5}\right) = 1\left(1 - \frac{2}{3}\right) = \frac{1}{3}\,\text{cm}.$

So to keep the image on the same film plane at $12\,\text{cm}$, the image formed by the lens alone must now be at $v_2 = 12 - \frac{1}{3} = \frac{35}{3}\,\text{cm}.$

3. Find new object distance

Using lens formula again: $\frac{1}{f} = \frac{1}{v_2} + \frac{1}{u_2}$ so $\frac{1}{u_2} = \frac{1}{f} - \frac{1}{v_2}.$

Substitute values: $\frac{1}{u_2} = \frac{7}{80} - \frac{3}{35}.$ Take LCM $560$: $\frac{1}{u_2} = \frac{49 - 48}{560} = \frac{1}{560}.$ Hence, $u_2 = 560\,\text{cm} = 5.6\,\text{m}.$

4. Match with options

Thus the object should be shifted to a distance $\boxed{5.6\,\text{m}}$ from the lens.

So the correct option is D.