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**Hint:**We will calculate the moment of inertia of the given solid rectangular door of mass M about the centre of mass of door using the formula ${I_{cm}} = \dfrac{{M{b^2}}}{{12}}$ and then we will use the parallel axis theorem to calculate the moment of inertia about the hinges of the door by using the formula: $I = {I_{cm}} + M{d^2}$ where I is the total moment of inertia and Icm is the moment of inertia about the centre of mass and d is the distance between the vertical axis and the centre of mass. After that, we will compare the given value to the obtained value to find the value of n.

**Complete step by step solution:**

We are given that a solid rectangular door is of uniform thickness.

Its width is given as 2 m.

The mass of the door is also provided which is 60 kg.

The distance between the centre of mass and the vertical axis of the door will be 1 m as we know that the width of the door is 2 m and hence the distance will be halved.

Now, the moment of inertia about the centre of the mass of the rectangular door will be given as: ${I_{cm}} = \dfrac{{M{b^2}}}{{12}}$

We have M = 60 kg and b = 2 m. on putting these values in the above equation, we get

$ \Rightarrow {I_{cm}} = \dfrac{{60 \times {2^2}}}{{12}} = \dfrac{{240}}{{12}} = 20kg{m^2}$

We have calculated the value of Icm and now we will calculate the value of the moment of inertia about the hinges of the door.

We have $I = {I_{cm}} + M{d^2}$, on putting values in this equation, we get

$ \Rightarrow I = 20 + 60 \times {1^2} = 20 + 60 = 80$

Therefore, the moment of inertia about the hinges of the door I = 80 kg$m^2$.

**Comparing it with the given value of moment of inertia which is 10n kg$m^2$, we get n = 8.**

**Note:**You may get confused while reading the given problem because both the moment of inertia is given i.e., the moment of inertia about the centre of mass and the moment of inertia about the vertical axis at a certain distance from the centre of mass of the body.

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