1. Find the set $A$

Since $2310 = 2\cdot 3\cdot 5\cdot 7\cdot 11,$ the set of all prime factors is $A=\{2,3,5,7,11\}.$ So, $|A|=5$.

2. Compute $f(x)$ for each $x\in A$

Given $f(x)=\left[\log_2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right].$ We evaluate one by one.

For $x=2$

$\left[\frac{2^3}{5}\right]=\left[\frac{8}{5}\right]=[1.6]=1.$ So, $x^2+\left[\frac{x^3}{5}\right]=4+1=5.$ Hence, $f(2)=\left[\log_2 5\right].$ Now, $2^2=4<5<8=2^3,$ so $\log_2 5\in(2,3)\implies \left[\log_2 5\right]=2.$ Thus, $f(2)=2.$

For $x=3$

$\left[\frac{3^3}{5}\right]=\left[\frac{27}{5}\right]=[5.4]=5.$ So, $x^2+\left[\frac{x^3}{5}\right]=9+5=14.$ Hence, $f(3)=\left[\log_2 14\right].$ Now, $2^3=8<14<16=2^4,$ so $\left[\log_2 14\right]=3.$ Thus, $f(3)=3.$

For $x=5$

$\left[\frac{5^3}{5}\right]=\left[\frac{125}{5}\right]=25.$ So, $x^2+\left[\frac{x^3}{5}\right]=25+25=50.$ Hence, $f(5)=\left[\log_2 50\right].$ Now, $2^5=32<50<64=2^6,$ so $\left[\log_2 50\right]=5.$ Thus, $f(5)=5.$

For $x=7$

$\left[\frac{7^3}{5}\right]=\left[\frac{343}{5}\right]=[68.6]=68.$ So, $x^2+\left[\frac{x^3}{5}\right]=49+68=117.$ Hence, $f(7)=\left[\log_2 117\right].$ Now, $2^6=64<117<128=2^7,$ so $\left[\log_2 117\right]=6.$ Thus, $f(7)=6.$

For $x=11$

$\left[\frac{11^3}{5}\right]=\left[\frac{1331}{5}\right]=[266.2]=266.$ So, $x^2+\left[\frac{x^3}{5}\right]=121+266=387.$ Hence, $f(11)=\left[\log_2 387\right].$ Now, $2^8=256<387<512=2^9,$ so $\left[\log_2 387\right]=8.$ Thus, $f(11)=8.$

3. Find the range of $f$

From above, $f(A)=\{2,3,5,6,8\}.$ All values are distinct, so the range has 5 elements.

Thus, $|\text{Range}(f)|=5.$

4. Count one-to-one functions from $A$ to the range of $f$

We need the number of injective functions from a 5-element set to another 5-element set.

Any one-to-one function from a 5-element set to a 5-element set is a bijection. Hence the number is $5! = 120.$

5. Final answer

$\boxed{120}$ So the correct option is B.