The co-ordinates of the centre of mass of the system as shown in figure are

The masses' (x, y) coordinates and masses can be expressed as m (0,0), m (L,0), and 3m (2L,0) (L, L). 

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3+{\mathrm{m}}_4{\mathrm{x}}_4}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+{\mathrm{m}}_4}$

${\mathrm{x}}_{\mathrm{cm}}=\frac{\mathrm{m}\times 0+\mathrm{m}\times \mathrm{L}+3\mathrm{m}\times 2\mathrm{L}+\mathrm{m}\times 0}{\mathrm{m}+\mathrm{m}+3\mathrm{m}+\mathrm{m}}=\frac{7\mathrm{mL}}{6\mathrm{m}}=1.33\mathrm{L}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2+{\mathrm{m}}_3{\mathrm{y}}_3+{\mathrm{m}}_4{\mathrm{y}}_4}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+{\mathrm{m}}_4}$

${\mathrm{y}}_{\mathrm{cm}}=\frac{0+0+0+\mathrm{mL}}{\mathrm{m}+\mathrm{m}+3\mathrm{m}+\mathrm{m}}=\frac{\mathrm{mL}}{6\mathrm{m}}=0.167\mathrm{L}$

Hence the correct answer is  (1.33L, 0.167L).