\[ B(x)=310 x^{2}-3720 x^{3}, 0 \leq x \leq 0.08 \] The maximum is obtained for a dosage of (Round to two decimal places as needed.)

Step 1 ：The function given is \(B(x) = 310x^2 - 3720x^3\) and we are asked to find the maximum value of this function in the interval \([0, 0.08]\).

Step 2 ：To find the maximum of a function, we need to find the derivative of the function, set it equal to zero, and solve for x. This will give us the critical points of the function.

Step 3 ：The derivative of \(B(x)\) is \(-11160x^2 + 620x\). Setting this equal to zero gives us the critical points \(x = 0\) and \(x = 1/18\).

Step 4 ：However, \(x = 1/18\) is approximately 0.0556, which is within the interval \([0, 0.08]\). Therefore, we need to check the value of \(B(x)\) at the endpoints of the interval, \(x = 0\) and \(x = 0.08\), to find the maximum value within the given interval.

Step 5 ：Evaluating \(B(x)\) at \(x = 0\) gives \(B(0) = 0\).

Step 6 ：Evaluating \(B(x)\) at \(x = 0.08\) gives \(B(0.08) = 0.0793599999999997\).

Step 7 ：Comparing these values, we find that the maximum value of \(B(x)\) in the interval \([0, 0.08]\) is obtained for a dosage of \(x = 0.08\).

Step 8 ：Final Answer: The maximum is obtained for a dosage of \(\boxed{0.08}\).