If F(0) = 0 , F(1)= 1....
Solve for n
F(n-1)< googol <F(n)
Use Binet's formula for the $\displaystyle n $-th Fibonacci number:
$\displaystyle F(n)=\frac{\varphi^n-(-1/\varphi)^n}{\sqrt{5}}$
where $\displaystyle \varphi=(1+\sqrt{5})/2 \approx 1.618.. $ is the Golden number
As $\displaystyle n$ will be very large $\displaystyle -(-1/\varphi)^n$ will be negligible so you need to find $\displaystyle n$ such that:
$\displaystyle \frac{\varphi^{n-1}}{\sqrt{5}}<\text{googol }<\frac{\varphi^{n}}{\sqrt{5}}$
CB
I agree with austinvishal on this one. Taking logs to base 10 (as we old-timers used to do all the time before they came up with these newfangled electronic thingies), we want to find $\displaystyle \frac{\log(\sqrt5\times10^{100})}{\log\varphi} \approx \frac{100.349485}{0.2089876}\approx480.17$. So we should take n = 481.