You are using floating point numbers. In most cases InverseFunction is used with symbolic numbers instead because it is a symbolic function. There are major differences between floating point and symbolic numbers.

Use Rationalize to convert most of your formula to symbolic numbers.

Rationalize@(0.203*Log[#] - 0.218*Log[0.015 - 0.203*#])/0.015
66.6667 (-(109/500) Log[3/200 - (203 #1)/1000] + (203 Log[#1])/1000)

This is enough to allow InverseFunction to work.

InverseFunction[66.6` (-(109/500) Log[3/200 - (203 #1)/1000] + (203 Log[#1])/1000)/0.015 &][201]

Remember though that you have to be careful with the precision you have. If we use Rationalize[#,0]& as work with a completely symbolic formula, it's possible to get a situation where there is no precision left in your result.

InverseFunction[200/3 (-(109/500) Log[3/200 - (203 #1)/1000] + (203 Log[#1])/1000) &][201.]

It is easy to make your calculations meaningless by losing all precision. I'm not sure how many digits of accuracy your coefficients have to begin with, but it's possible that InverseFunction might not be a very stable operation in this case.

The difference in value is probably floating point error. If you're comfortable knowing the result is close to 0, then the result might be fine. How much precision are you look for in your result? You may be able to get it using arbitrary precision arithmetic.

http://reference.wolfram.com/language/tutorial/ArbitraryPrecisionNumbers.html

If InverseFunction isn't working out at all, then some serious numerical work will have to be done to get a good answer. Typically, computing the inverse of a function is viewed as a root finding problem. You may want to experiment with different Methods for FindRoot. The three statements below are essentially equivalent.

NSolve[f[x] == b,x]

x = InverseFunction[f][b]

FindRoot[f[x]-b, {x,0}]

I may have miscopied the code. The suggestion however is to try computing the values symbolically in some way. Or you may want to investigate why this isn't numerically stable.

Rationalize[(0.203*Log[#] - 0.218*Log[0.015 - 0.203*#])/0.015]
200/3 (-(109/500) Log[3/200 - (203 #1)/1000] + (203 Log[#1])/1000)

InverseFunction[200/3 (-(109/500) Log[3/200 - (203 #1)/1000] + (203 Log[#1])/1000) &][201]

This gives back a Root expression, which is just a complicated way of describing a number. You can run N on it to try returning a decimal approximation. You may need to use arbitrary precision computing however.

I think I was not clear in my previous post.

I tried to find the inverse function using 2 methods for the same expression. Directly solving using InverseFunction gave me a result which is 0.0101108 ; while solving it first by using Rationalize and then solving it using N gave me a result which is 4.26674*10^9.

Please note that there are many orders of magnitude difference between the both but not a floating point error.