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**Hint:-**In this question the sample of metal weights in air is 210g which is the actual weight of the metal, we need to form two equations by using the formula$D = \dfrac{m}{v}$; D = density; m = mass; v = volume and then use relative density to find the density of the metal . Here the relative density would be the ratio of two densities, the density of mass upon the density of water.

**Complete step-by-step solution:-**

Write down the weight of the metal in relation with density and volume.

$D = \dfrac{m}{v}$ ;

Take volume to LHS

$v \times D = m$;

For material the equation becomes

${v_m} \times {\rho _m} = {m_m}$; …(${\rho _m}$= D = Density of metal)

Put the given value,

${v_m} \times {\rho _m} = 0.21$;

To get the density for the material we have to subtract the density of water from the density of material,

$0.21 - {v_w} \times {\rho _w} = {m_w}$; …(${\rho _w}$= density of water)

Put the value of mass of material in water,

$ - {v_w} \times {\rho _w} = 0.18 - 0.21$;

Solve, negative sign will cancel each other out,

${v_w} \times {\rho _w} = 0.3$;

Now here the relative density would be the density of material upon the density of water,

$\dfrac{{{\rho _m}}}{{{\rho _w}}} = \dfrac{{0.21}}{{0.3}}$ ;

Solve,

\[\dfrac{{{\rho _m}}}{{{\rho _w}}} = 0.7\];

Now, take the density of water to be $1gm/c{m^3}$

${\rho _m} = 0.7 \times {\rho _w}$;

Put the value in the above equation,

${\rho _m} = 0.7 \times 1$;

The density of the metal,

${\rho _m} = 7gm/c{m^3}$;

**Final Answer:**Option “2” is correct. The density of metal is$7gm/c{m^3}$.

**Note:-**Here we have to find the relative density. The relative density is the ratio between the two densities. To find the density of the metal we have to first find the density of material in the water and then find the relative density. Take the density of the water to be $1gm/c{m^3}$.

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