1.333333333333

Since $\sqrt [ 3 ]{ x } +y+z$ , $x+\sqrt [ 3 ]{ y } +z$ and $x+y+\sqrt [ 3 ]{ z }$ give us integer answer.

$\Rightarrow$ Let $\sqrt [3]{x}=a$ , $\sqrt [3]{y}=b$ and $\sqrt [3]{z}=c$ , we can rewrite the question to

$a^3+b^3+c^3 = 2196$ --------(1)

$a+b^3+c^3 = 2076$ -------(2)

$a^3+b+c^3 = 1860$ -------(3)

$a^3+b^3+c =480$ -------(4)

(1) - (2) , we get $a^3-a=120$

Factorizes both size of the equation, $a(a-1)(a+1) = 2^3 \times 3 \times 5\\a(a+1)(a-1)=4 \times 5 \times 6 \\\boxed{a=5}$

We can find the value of $b$ and $c$ by the same method.

(1) - (3) , we get $b^3-b=336$

Factorizes both size of the equation, $b(b-1)(b+1) = 2^4 \times 3 \times 7\\b(b+1)(b-1)=6 \times 7 \times 8 \\\boxed{b=7}$

(1) - (4) , we get $c^3-c=1716$

Factorizes both size of the equation, $c(c-1)(c+1) = 2^2 \times 3 \times 11 \times 13\\c(c+1)(c-1)=11 \times 12 \times 13 \\\boxed{c=12}$

Therefore, ${x}^{\frac {2}{3} }+{ y }^{ \frac {2}{3}}+{ z }^{ \frac{2}{3}}\\=a^2+b^2+c^2\\=5^2+7^2+12^2\\=25+49+144\\=\boxed{218}$