1. Given data

• Length of solid cylinder: $L = 80\text{ cm} = 0.8\text{ m}$
• Radius: $R = 20\text{ cm} = 0.2\text{ m}$
• Moment of inertia about axis $CD$: $I_{CD} = 2.7\,\text{kg m}^2$

From the statement "axis $CD$ parallel to $AB$ as shown", this is the standard case where $AB$ is the central longitudinal axis of the cylinder and $CD$ is a parallel axis touching the surface. Hence the perpendicular distance between the two axes is

$d = R = 0.2\text{ m}.$

2. Moment of inertia about the central longitudinal axis

For a solid cylinder about its own axis,

$I_{AB} = \frac{1}{2}MR^2.$

Using parallel axis theorem,

$I_{CD} = I_{AB} + Md^2.$

So,

$2.7 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2.$

Substitute $R = 0.2$ m:

$2.7 = \frac{3}{2}M(0.2)^2$ $2.7 = \frac{3}{2}M(0.04)$ $2.7 = 0.06M$

Therefore,

$M = \frac{2.7}{0.06} = 45\text{ kg}.$

3. Volume of the cylinder

$V = \pi R^2 L$ $V = \pi (0.2)^2(0.8)$ $V = \pi (0.04)(0.8)$ $V = 0.032\pi\text{ m}^3.$

4. Density of the material

$\rho = \frac{M}{V} = \frac{45}{0.032\pi}.$

Using $\pi \approx 3.14$,

$\rho \approx \frac{45}{0.10048} \approx 447.9\,\text{kg/m}^3.$

So,

$\boxed{\rho \approx 4.48 \times 10^2\,\text{kg/m}^3}.$

5. Check against options

The calculated density $\approx 4.48 \times 10^2\,\text{kg/m}^3$ does not match any of the given options.

6. Comparison with stored answer

Stored correct answer is D: $1.49 \times 10^2\,\text{kg/m}^3$.

But our calculation gives

$\boxed{4.48 \times 10^2\,\text{kg/m}^3}$

so I do not agree with the stored answer. The likely issue is either:

• the figure implies a different axis separation than stated, or
• the given moment of inertia value has a typo.