Correct Answer: $4:1$

Solution : Given: $(x^{3}-y^{3}):( x^{2}+xy+y^{2})=5:1$ and $( x^{2}-y^{2}):(x-y)=7:1$
Now, $\frac{x^{3}-y^{3}}{ x^{2}+xy+y^{2}}=\frac{5}{1}$
Using the algebraic identity, $a^3-b^3=(a-b)(a^2+ab+b^2)$
⇒ $\frac{(x-y)(x^{2}+xy+y^{2})}{ x^{2}+xy+y^{2}}=\frac{5}{1}$
⇒ $x-y=5$ ......................... (1)
Again, $\frac{x^{2}-y^{2}}{ x-y}=\frac{7}{1}$
Using the algebraic identity $a^2-b^2=(a-b)(a+b)$
⇒$\frac{(x-y)(x+y)}{ x-y}=\frac{7}{1}$
⇒ $x+y=7$ ........................ (2)
Solving both equations, we get:
⇒ $x= 6, y=1$
Thus, $2x:3y$ = $2×6:3×1$ = $12:3$ = $4:1$
Hence, the correct answer is $4:1$.