Atoms of metals x, y, and z form face-centred cubic (fcc) unit cell of edge length $$L_x$$, body-centred cubic (bcc) unit cell of edge length $$L_y$$, and simple cubic unit cell of edge length $$L_z$$, respectively.

If $$r_z = \frac{\sqrt{3}}{2}r_y$$; $$r_y = \frac{8}{\sqrt{3}}r_x$$; $$M_z = \frac{3}{2}M_y$$ and $$M_z = 3M_x$$, then the correct statement(s) is (are)

[Given: $$M_x$$, $$M_y$$, and $$M_z$$ are molar masses of metals x, y, and z, respectively. $$r_x$$, $$r_y$$, and $$r_z$$ are atomic radii of metals x, y, and z, respectively.]

For every cubic lattice the density, edge length and packing efficiency depend on three basic data: number of atoms per unit cell ($$Z$$), atomic radius ($$r$$) and molar mass ($$M$$).
We tabulate the standard relations first:

Face-centred cubic (fcc, metal $$x$$):
  • $$Z_x = 4$$     • contact along a face diagonal ⇒ $$\sqrt{2}\,L_x = 4\,r_x$$ ⇒ $$r_x = \dfrac{L_x}{2\sqrt{2}}$$.
Body-centred cubic (bcc, metal $$y$$):
  • $$Z_y = 2$$     • contact along a body diagonal ⇒ $$\sqrt{3}\,L_y = 4\,r_y$$ ⇒ $$r_y = \dfrac{\sqrt{3}}{4}\,L_y$$.
Simple cubic (sc, metal $$z$$):
  • $$Z_z = 1$$     • contact along an edge ⇒ $$L_z = 2\,r_z$$ ⇒ $$r_z = \dfrac{L_z}{2}$$.

The question gives the inter-relations

$$r_z = \frac{\sqrt{3}}{2}\,r_y,\qquad r_y = \frac{8}{\sqrt{3}}\,r_x,\qquad M_z = \frac{3}{2}\,M_y,\qquad M_z = 3\,M_x.$$

From the last two equalities

$$M_x = \frac{1}{3}M_z,\qquad M_y = \frac{2}{3}M_z \;\;\Longrightarrow\;\; \frac{M_x}{M_y} = \frac12.$$

1. Comparison of edge lengths

Using $$r_y = \dfrac{8}{\sqrt{3}}\,r_x$$ in $$L_y = \dfrac{4\,r_y}{\sqrt{3}}$$:
$$L_y = \frac{4}{\sqrt{3}}\left(\frac{8}{\sqrt{3}}\,r_x\right) = \frac{32}{3}\,r_x.$$
Using $$r_z = \frac{\sqrt{3}}{2}\,r_y = \frac{\sqrt{3}}{2}\left(\frac{8}{\sqrt{3}}\,r_x\right)=4\,r_x$$ in $$L_z=2\,r_z$$:
$$L_z = 8\,r_x.$$
For fcc, $$L_x = 2\sqrt{2}\,r_x.$$

Numerically (in units of $$r_x$$):
$$L_x = 2.828\,r_x,\qquad L_y = 10.667\,r_x,\qquad L_z = 8\,r_x.$$ Therefore $$L_y \gt L_z \gt L_x$$ and hence Option B ($$L_y \gt L_z$$) is correct, while Option C ($$L_x \gt L_y$$) is incorrect.

2. Packing efficiencies

Standard values are: fcc = 74 %, bcc = 68 %, sc = 52.4 %.
Thus$$\text{fcc (}x\text{)} \gt \text{bcc (}y\text{)} \gt \text{sc (}z\text{)},$$so Option A is correct.

3. Density comparison of $$x$$ and $$y$$

For any cubic lattice $$\rho = \dfrac{Z\,M}{N_A\,L^3}.$$ Taking the ratio $$ \frac{\rho_x}{\rho_y} = \frac{Z_x M_x L_y^{\,3}}{Z_y M_y L_x^{\,3}} = \left(\frac{4}{2}\right) \left(\frac{M_x}{M_y}\right) \left(\frac{L_y}{L_x}\right)^{3} = 2\left(\frac12\right)\left(\frac{L_y}{L_x}\right)^{3} = \left(\frac{L_y}{L_x}\right)^{3}. $$ We already have $$\dfrac{L_y}{L_x} = \dfrac{32/3}{2\sqrt{2}} = \dfrac{16}{3\sqrt{2}} \approx 3.772,$$ so $$\frac{\rho_x}{\rho_y} \approx 3.772^{3} \approx 53.7 \gt 1.$$ Hence $$\rho_x \gt \rho_y,$$ making Option D correct.

Final verdict:
Correct statements: Option A, Option B, Option D.